Integration by parts tute0061   update 2015-02-14
Zeta function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12, 13, 14, 15, 16, 17
Gamma function: 01, 02, 03, 04, 05, 06, 07, 08, 09, 10, 11, 12
jw1.nwnu.edu.cn integration
Integral Test, Euler's formula, Zeta video list, how to download video




<a name="Gamma_Indx"> 
2015-01-01-15-18 start 
On 
2014-03-25-20-36 Liu,Hsinhan access
http://www.youtube.com/playlist?list=PL3E4136E122545FBE
find and download next 12 video files. 

Gamma Function 
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
<a name="GammaIndex"> 
Gamma Function - Part 1 - Functional Equation
http://www.youtube.com/watch?v=2iBNo4j3vRo
Gamma Function - Part 2 - Gauss Representation
http://www.youtube.com/watch?v=a_Dlx7TTjkI
Gamma Function - Part 3 - Weierstrass Representation
http://www.youtube.com/watch?v=d9Oz62Ioue0
Gamma Function - Part 4 - Relationship to Sine
http://www.youtube.com/watch?v=W3ulxj3s90U
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
http://www.youtube.com/watch?v=_vwqsJNKY-c
Gamma Function - Part 6 - Stirling's Approximation
http://www.youtube.com/watch?v=MuAb2dnPD0Q
An intuitive derivation of Stirling’s formula
http://www.rowan.edu/open/depts/math/osler/Stirlings_formula.pdf

Gamma Function - Part 7 - Euler Integral I
http://www.youtube.com/watch?v=VF7ud3Al6d8
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
http://www.youtube.com/watch?v=h4wQ_F1s1UI
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
http://www.youtube.com/watch?v=H0kT-EKbUzM
Gamma Function - Part 10 - Beta Function
http://www.youtube.com/watch?v=Korx_G7eySQ
Gamma Function - Part 11 - Legendre Duplication Formula
http://www.youtube.com/watch?v=yu9NeDXalpA
Gamma Function - Part 12 - Relation to Zeta Function
http://www.youtube.com/watch?v=jQAPlsNY_P0
Above red link has complete lecture. Blue not.
Gauss multiplication formula half way, not done!
https://proofwiki.org/wiki/Gauss_Multiplication_Formula
Γ(z) has a simple pole with residue (-1)n/n! 
at z=-n for n = 0, 1, 2, ...
http://www.math.leidenuniv.nl/~evertse/ant13-8.pdf
2015-01-01-15-38 stop

<a name="ZetaIndex"> 
2014-10-29-07-56
https://www.youtube.com/playlist?list=PL32446FDD4DA932C9
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence" 
http://www.youtube.com/watch?v=ZlYfEqdlhk0
"Zeta Function - Part 2 - Euler Product Representation" 
http://www.youtube.com/watch?v=I3qSCWNXZKg
"Zeta Function - Part 3 - Euler Product (easy)" 
http://www.youtube.com/watch?v=TDdGisWD5OU
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers" 
http://www.youtube.com/watch?v=SKa7b-3C32A
"Zeta Function - Part 5 - Prime Zeta Function" 
http://www.youtube.com/watch?v=3eN9tQX3JJ4
"Zeta Function - Part 6 - The Prime Counting Function" 
http://www.youtube.com/watch?v=U16_KTTKtb0
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem" 
http://www.youtube.com/watch?v=GeKDmoYHiAk
"Zeta Function - Part 8 - Zeta of 2n - Part 1" 
http://www.youtube.com/watch?v=axQqExF7NsU
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2" 
http://www.youtube.com/watch?v=XHQ0OzqTjd0 9/17
"Zeta Function - Part 8 - Zeta of 2n - Part 3" 
http://www.youtube.com/watch?v=1f24RZfP6m8 10/17
"Zeta Function - Part 9 - Relation to Gamma Function" 
http://www.youtube.com/watch?v=UEZ4ClCdog8 11/17
"Zeta Function - Part 10 - Jacobi Theta Function" 
http://www.youtube.com/watch?v=-GQFljOVZ7I 12/17
"Zeta Function - Part 11 - Riemann Functional Equation I" 
http://www.youtube.com/watch?v=K6L4Ez4ZVZc 13/17
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II" 
http://www.youtube.com/watch?v=TnRnlJBecRg 14/17
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function" 
http://www.youtube.com/watch?v=G-fqe3BkBnE 15/17
"Zeta Function - Part 14 - Riemann Xi Function" 
http://www.youtube.com/watch?v=QfDbF_qlp58 16/17
"Sine Function Product Formula (Hadamard Factorization Theorem)"
https://www.youtube.com/watch?v=h3Hcioh2F9I 17/17
"What is a function? Why 1+2+3+4+5+.... not equals -1/12 = Zeta(-1)" 
http://www.youtube.com/watch?v=wt6ngy6pDws This video is easy. skipped a312190901
<a name="a40103a"> Integration by parts derivation
jw1.nwnu.edu.cn integration
Below is Liu,Hsinhan derivation, 
may contain error. Please verify. 
2015-01-03-15-05 start 
This file tute0061.htm is study notes for 
Integration by parts.
Next file tute0062.htm is study notes for 
Gamma Function, lectured by MrYouMath.
Before study Gamma Function, need review 
integration by parts. 
Assume f(x) is a function of one independent 
variable x. For example 
 f(x)=x2 --- eq.LA01 
or 
 f(x)=sin(x) --- eq.LA02 
<a name="a40103b">
Differentiation of f(x) with respect to x is 
 
df(x)

dx
define
═══
lim
h→0
f(x+h)-f(x)

(x+h)-x
=
lim
h→0
f(x+h)-f(x)

h
=
f '(x)
--- eq.LA03
h→0 : h approach to zero, but h not equal to zero. h/h= 1 is possible.
eq.LA03 mean equation LiuHH sequence A (ABCDE ... ) ; a401071541
width of above equation a401031520
<a name="a40103c">
2015-01-03-15-21 here
In eq.LA03, left equality is defined. 
In 'df(x)' and in 'dx', 'd' mean "take a very 
small section of" . 
'f(x+h)-f(x)' is a small section of f(x). 
lim[h→0] indicate "very small section". 
'h' approach to zero, but h NOT= 0, such that 
division is possible. 
"f'(x)" is a short write for next longer line. 
lim[h→0]{[f(x+h)-f(x)]/h}

<a name="a40103d">
Above function is f(x). What if function is 
a product of two? d[f(x)*g(x)]/dx ? 
Follow eq.LA03, write eq.LA04 below. 
<a name="a40103e">
 
d[f(x)*g(x)]

dx
=
lim
h→0
f(x+h)*g(x+h)-f(x)*g(x)

h
=
lim
h→0
f(x+h)*g(x+h)-f(x)*g(x) +0

h
--- eq.LA04
 
d[f(x)*g(x)]

dx
=
lim
h→0
f(x+h)*g(x+h)-f(x)*g(x) +f(x+h)*g(x) -f(x+h)*g(x)

h
--- eq.LA05
 
d[f(x)*g(x)]

dx
=
lim
h→0
f(x+h)*g(x+h) -f(x+h)*g(x)     -f(x)*g(x) +f(x+h)*g(x)

h
--- eq.LA06
 
d[f(x)*g(x)]

dx
=
lim
h→0
f(x+h)*[g(x+h)-g(x)]

h
lim
h→0
+g(x)[+f(x+h)-f(x)]

h
--- eq.LA07
width of above equation a401031557

lim[h→0]f(x+h) is f(x) and lim[h→0]{[g(x+h)-g(x)]/h} is g '(x)
<a name="a40103f">
eq.LA08 is differentiation product rule.
 
d[f(x)*g(x)]

dx
  =     f(x) *
dg(x)

dx
  +     g(x) *
df(x)

dx
--- eq.LA08
In eq.LA08 cancel denominator dx get // dx NOT= 0
d[f(x)*g(x)] = d[f(x)]*g(x) + f(x)*d[g(x)] --- eq.LA09

width of above equation a401031603
<a name="a40103g">
104,01,03,16,42 here 
Next see integration. 
 
x=b
x=a
dx = b-a  ;
x=b
x=a
h(x)dx =
x=b
x=a
dH(x) = H(b)-H(a)
--- eq.LA10
Here dH(x)/dx = h(x) ; if H(x)=x2, then h(x)=2x ; if H(x)=log(x), then h(x)=1/x ;
width of above equation a401031650

<a name="a40103h">
Assume dF(x)/dx = f(x) and dG(x)/dx = g(x) , assume H(x)=F(x)*G(x)
next is Integration by parts

 
F(x)G(x)
x=b
x=a
=
x=b
x=a
d[F(x)G(x)] eq.LA09
═══
x=b
x=a
G(x)*dF(x) + F(x)*dG(x)
--- eq.LA11
 
F(x)G(x)
x=b
x=a
=
x=b
x=a
G(x)*f(x)dx + F(x)*g(x)dx
--- eq.LA12
 
F(x)G(x)
x=b
x=a
=
x=b
x=a
G(x)f(x)dx +
x=b
x=a
F(x)g(x)dx
--- eq.LA13
<a name="a40103i"> Goto Integration by parts derivation begin ; priority table 1,2,3
Black term harder to solve. red term be zero in some case. Blue term easier.
 
x=b
x=a
F(x)g(x)dx =
F(x)G(x)
x=b
x=a
x=b
x=a
G(x)f(x)dx
--- eq.LA14
 
x=b
x=a
U(x)v(x)dx =
U(x)V(x)
x=b
x=a
x=b
x=a
V(x)u(x)dx
--- eq.LA15
Integration by parts Some textbook use u(x), v(x).
width of above equation a401031734

Above is Liu,Hsinhan derivation, may contain error. Please verify.
2015-01-03-17-36
<a name="a40103j">
2015-01-03-22-18 start 
eq.LA03 is function f(x) differentiation 
definition equation. Independent variable 
x change from x to x+h, the difference is 
(x+h)-x. At same time function value change 
is f(x+h)-f(x). The ratio of f(x+h)-f(x) to 
(x+h)-x is defined to be derivative of f(x).
f(x+h)-f(x) is a line segment on y axis 
where y=f(x) . (x+h)-x is a line segment on 
x axis. If h is finite value, then 
[f(x+h)-f(x)]/[(x+h)-x] is secant line slope 
not tangent line slope. Take limit let h→0 
then limit[h→0]{[f(x+h)-f(x)]/[(x+h)-x]}
approach tangent line slope. 
In eq.LA03 f'(x) is a short write of df(x)/dx

<a name="a40103k">
eq.LA04 is differentiation of f(x)*g(x)
eq.LA04 apply eq.LA03 to f(x)*g(x)
eq.LA04 add a zero. zero is red term in eq.LA05 
Redistribute eq.LA05 numerator terms to eq.LA06
and re-group to eq.LA07. Apply  eq.LA03 definition 
twice get eq.LA08 
eq.LA08 is differentiation product rule.
2015-01-03-22-58 

<a name="a40103l">
eq.LA11 left equality is integration of [F(x)G(x)]  
and evaluate at x=a and x=b.
eq.LA11 right equality apply eq.LA09 
eq.LA11 to eq.LA12 apply dH(x)/dx = h(x) 
if H(x)=x2, then h(x)=2x ; 
if H(x)=log(x), then h(x)=1/x ;
eq.LA12 to eq.LA13 write integration sign 
to two terms. 
<a name="a40103m">
eq.LA13 to eq.LA14 write whole equation in 
Integration by parts form. 
In eq.LA14 
Left side black term harder to solve. 
red term be zero in some case. not all case.
Blue term easier to solve..  
2015-01-03-23-20 stop

<a name="a40104a">
2015-01-04-09-30 start 
Let u(x) and v(x) be two functions. 
Let U(x) be antiderivative of u(x) . 
Let V(x) be antiderivative of v(x) . 

<a name="a40104b">
Example 1:
If u(x)=1, U(x)=x+C, C be a constant.
because 
dU(x)/dx = d(x+C)/dx = dx/dx + dC/dx 
         = 1+0 = 1 = u(x) --- eq.LA16 
If function f(x)=C=constant, then 
f'(x)=limit[h→0]{[f(x+h)-f(x)]/[(x+h)-x]}
f'(x)=limit[h→0]{[C-C]/[(x+h)-x]}=0 --- eq.LA17 

<a name="a40104c">
Example 2:
If u(x)=x, U(x)=C+[x2]/2, C be a constant.
because 
dU(x)/dx = d(C+[x2]/2)/dx = dC/dx + d([x2]/2)/dx 
         = 0 + 2*x/2 = x = u(x) --- eq.LA18 

<a name="a40104d">
Example 3:
If u(x)=xn, U(x)=C+[xn+1]/[n+1] .
because 
dU(x)/dx = d(C+[xn+1]/[n+1])/dx 
         = dC/dx + d([xn+1]/[n+1])/dx 
         = 0 + [n+1]*xn/[n+1] = xn = u(x) --- eq.LA19 

<a name="a40104e">
This equation 
 dU(x)/dx = u(x)  --- eq.LA20 
can be written as 
 dU(x) = u(x)*dx  --- eq.LA21 
If u(x)=x2/3 then U(x) MUST BE x5/3 
'dx' contribute to x1=x3/3 'x' HAS dimension. 
'd'=tiny amount, 'd' has no dimension.
x2/3*x = x2/3*x3/3 = x5/3 
Because in eq.LA21 we can drop 'd' for 
dimension analysis. 'd' indicate tiny amount.
<a name="a40104f">
After drop 'd', eq.LA21 is 
  U(x) = u(x)* x  --- eq.LA22 
If u(x)=K*x2/3 is a force representation 
K is proportional coefficient. 
If x is distance traveled, then 
u(x)* x is work done. u(x)* x is K*x2/3*x .
u(x)* x is K*x5/3 . Remember x2/3*x = x5/3 
U(x) physics dimension must absorb x dimension 
to change force u(x) to U(x) energy (work). 
2015-01-04-10-08

<a name="a40104g">
Example 4:
If u(x)=ex, U(x)=C+ex .
because 
dU(x)/dx = d(C+ex)/dx 
         = dC/dx + d(ex)/dx 
         = 0 + ex = ex = u(x) --- eq.LA22 

<a name="a40104h">
Example 5:
If u(x)=1/x, U(x)=C+log(x) .
because 
dU(x)/dx = d(C+log(x))/dx 
         = dC/dx + d(log(x))/dx 
         = 0 + [1/x] = 1/x = u(x) --- eq.LA23 

<a name="a40104i">
Example 6:
If u(x)=cos(x), U(x)=C+sin(x) .
because 
dU(x)/dx = d(C+sin(x))/dx 
         = dC/dx + d(sin(x))/dx 
         = 0 + cos(x) = cos(x) = u(x) --- eq.LA24 

<a name="a40104j">
Example 7:
If u(x)=sin(x), U(x)=C-cos(x) .
because 
dU(x)/dx = d(C-cos(x))/dx 
         = dC/dx - d(cos(x))/dx 
         = 0 - [-sin(x)] = sin(x) = u(x) --- eq.LA25 

Above is seven antiderivative examples 

<a name="a40104k">
Let u(x) and v(x) be two functions. 
Let U(x) be antiderivative of u(x) . 
Let V(x) be antiderivative of v(x) . 

Next see Integration by parts equation eq.LA15 
the role for u(x),v(x) and U(x),V(x)
2015-01-04-10-23
<a name="a40104l"> Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts
 
x=b
x=a
U(x)v(x)dx =
U(x)V(x)
x=b
x=a
x=b
x=a
V(x)u(x)dx
--- eq.LA15
width of above equation a401040925
<a name="a40104m">
2015-01-04-10-24 here 
In eq.LA15, two questions.
What is the difference between U(x) and v(x)?

If given U(x)v(x)=xn*ex, choose which one to 
be U(x)? Choose xn to be U(x), ex to be v(x)? 
or choose ex to be U(x), choose xn to be v(x)?

In eq.LA15 left side has U(x)v(x), at right 
side, U(x) and v(x) have different 'fate'.
<a name="a40104n">
Left side U(x) will be differentiated at 
right side and become u(x)dx This u(x) is 
easier than U(x). 
Left side v(x) will be integrated at right 
side and become V(x) Right side V(x) is not 
too hard than left side v(x). 

<a name="a40104o">
If given U(x)v(x)=xn*ex. 
Correctly choose xn to be U(x), ex to be v(x) 
then right side is 
xn*ex▎[x=a,x=b] -∫[x=a,x=b]ex*n*xn-1*dx
x power change from xn to xn-1 which is simpler.

<a name="a40104p">
Wrongly choose ex to be U(x), choose xn to 
be v(x) then right side is 
xn+1*ex/(n+1)▎[x=a,x=b] 
-∫[x=a,x=b][xn+1/(n+1)]*ex*dx
x power change from xn to xn+1 which is harder.

<a name="a40104q">
eq.LA15 left side (black term) is harder to 
solve integration problem. 
eq.LA15 right side U(x)V(x)|[x=a,x=b] (red term) 
is boundary terms. Because set x=a and x=b at 
two integration ends, it is called boundary terms.
eq.LA15 right side blue color new integration , 
▬∫[x=a,x=b]{V(x)u(x)dx }
this is an integration problem easier to solve. 
2015-01-04-12-33 stop 

<a name="a40104r">
2015-01-04-14-29 start 
Liu,Hsinhan major in mechanical engineering.
Not major in mathematics. Above are LiuHH's 
own words. LiuHH notes may contain error. 
Reader must suspect each step, reader must 
exam each word. 
Below is mathematics teacher's lecture. 
This is a Chinese video. URL valid in 2013. 
<a name="a40104s">
2013-09-16-16-20  download [下載] hxr.rar
主講教師洪學仁講師教學錄像(第四章第三節 分部積分法) [播放][下載]
http://jw1.nwnu.edu.cn/jpkc/wdxy/gdsxI/jxlx_hxr.asp
http://jw1.nwnu.edu.cn/jpkc/wdxy/gdsxI/jxlx/hxr.rar
LiuHH record ten examples in the lecture. 
Below write eq.LA15 as ∫udv=uv-∫vdu 

<a name="a40104t">
Example 1. if given 
u(x)v(x)=xn*sin(kx) or =xn*cos(kx) 
Differentiation will reduce xn power, but 
differentiation cannot simplify sin(kx).
In eq.LA15 U(x) will be differentiated at 
right side as u(x)dx .
choose U(x)=xn=x to reduce power n.
<a name="a40104t1">
Blue is chosen to reduce power n. Red is final answer.
∫p over '=' indicate here apply Integration by parts equation eq.LA15 //a401071750
 
x*cos(x)*dx =
x*d[sin(x)] ∫p
=
x*sin(x)
sin(x)*dx
--- eq.LA26
 
x*cos(x)*dx = x*sin(x) +
d[cos(x)] = x*sin(x) + cos(x) + C
--- eq.LA27
width of above equation a401041459
<a name="a40104u">
2015-01-04-15-13 here
Example 2. if given 
u(x)v(x)=x2*sin(3x)*dx
choose U(x)=x2 to reduce power 2
Blue U(x) is chosen to differentiate to u(x) . Red is final answer. <a name="a40104u1">
Purple v(x) is chosen to integrate to V(x). Present purple color only in Example 2.
 
x2*sin(3x)*dx =
x2*sin(3x)*d(3x)

3
=
x2*d[-cos(3x)]

3
--- eq.LA28
 
∫p
=
x2*[-cos(3x)]

3
[-cos(3x)]*d[x2]

3
=
-x2*cos(3x)

3
-cos(3x)*2*x*dx

3
--- eq.LA29
 
=
-x2*cos(3x)

3
+
+cos(3x)*2*x*dx

3
=
-x2*cos(3x)

3
+
2*x*cos(3x)*d(3*x)

3*3
---a40104u2
--- eq.LA30
 
=
-x2*cos(3x)

3
+
2*x*d[sin(3x)]

9
∫p
=
-x2*cos(3x)

3
+
2x*sin(3x)

9
-
2*sin(3x)*dx

9
--- eq.LA31
 
x2*sin(3x)*dx =
-x2*cos(3x)

3
+
2x*sin(3x)

9
-
2*sin(3x)*d(3x)

9*3
---a40104u3
--- eq.LA32
 
x2*sin(3x)*dx =
-x2*cos(3x)

3
+
2x*sin(3x)

9
+
2*cos(3x)

27
+ C
--- eq.LA33
width of above equation a401041622
2015-01-04-16-22 stop
<a name="a40104v">
2015-01-04-18-02 start
Example 3. if given 
u(x)v(x)=xn*ek*x ; n=1 and k=1 
Differentiation will reduce xn power, but 
differentiation cannot simplify ek*x.
In eq.LA15 U(x) will be differentiated at 
right side as u(x)dx .
choose U(x)=xn=x to reduce power n.
Blue is chosen to reduce power n. Red is final answer. <a name="a40104v1">
 
x*ex*dx =
x*d[ex] ∫p
=
x*ex
ex*dx
--- eq.LA34
 
x*ex*dx = x*ex -
d[ex] = x*ex - ex + C
--- eq.LA35
width of above equation a401041816
<a name="a40104w">
2015-01-04-18-18
Example 4. if given 
u(x)v(x)=x2*e3*x 
Differentiation will reduce x2 power, but 
differentiation cannot simplify e3*x.
In eq.LA15 U(x) will be differentiated at 
right side as u(x)dx .
choose U(x)=x2 to reduce power 2.
Blue is chosen to reduce power n. Red is final answer. <a name="a40104w1">
 
x2*[e3*x]*dx =
x2*[e3*x]*d(3*x)

3
=
x2*d[e3*x]

3
--- eq.LA36
 
∫p
=
x2*e3*x

3
[e3*x]*d(x2)

3
=
x2*e3*x

3
[e3*x]*2x*dx

3
--- eq.LA37
 
=
x2*e3*x

3
2x*[e3*x]*dx

3
=
x2*e3*x

3
2x*[e3*x]*d(3*x)

3*3
a40104w2
--- eq.LA38
 
=
x2*e3*x

3
2x*d[e3*x]

3*3
∫p
=
x2*e3*x

3
2x*e3*x

9
+
2*[e3*x]*dx

3*3
--- eq.LA39
 
=
3*x2*e3*x-2x*e3*x

9
+
2*[e3*x]*d(3*x)

3*3*3
=
3*x2*e3*x-2x*e3*x

9
+
2*d[e3*x]

3*3*3
a40104w3
--- eq.LA40
 
x2*[e3*x]*dx   =  
x2*e3*x

3
2x*e3*x

9
+
2*e3*x

27
+ C
--- eq.LA41
width of above equation a401041907
2015-01-04-19-08 stop
<a name="a40104x">
2015-01-04-21-00 start 
Example 1,2 is xn and sin(x) choose xn as U(x).
Example 3,4 is xn and e3*x , choose xn as U(x).
Example 5,6 below is xn and log(x) 
choose xn as U(x) ? dxn/dx=n*xn-1
then   log(x) antiderivative is x*log(x)-x
choose log(x) as U(x) ? dlog(x)/dx=1/x 
then   xn antiderivative is xn+1/(n+1)
Consider log(x) antiderivative and derivative 
it is better to choose log(x) as U(x) !
Because 1/x is easier than x*log(x)-x

<a name="a40104y">
Example 5. if given 
∫u(x)v(x)dx=∫x2/3*log(x)*dx
choose log(x) as U(x)
Blue is chosen to differentiate. Red is final answer. <a name="a40104y1">
 
x2/3*log(x)*dx =
log(x)*dx5/3

5/3
∫p
=
log(x)*x5/3

5/3
x5/3*d[log(x)]

5/3
--- eq.LA42
 
=
log(x)*x5/3

5/3
x5/3*dx

x*5/3
=
log(x)*x5/3

5/3
x2/3*dx

5/3
--- eq.LA43
 
=
log(x)*x5/3

5/3
dx5/3

(5/3)*(5/3)
=
log(x)*x5/3

5/3
x5/3

(5/3)*(5/3)
+ C
a40104y2
--- eq.LA44
 
       
x2/3*log(x)*dx =    
3*log(x)*x5/3

5
9*x5/3

25
+ C
       
--- eq.LA45
width of above equation a401042149
<a name="a40104z">
2015-01-04-21-50
Example 6. if given 
∫u(x)v(x)dx=∫log(1+x2)*dx
ONLY choose log(1+x2) as U(x)
The other 'function' is constant 1. If choose 
1 as U(x), d(constant)/dx=0. Get nothing. 
Blue is chosen to differentiate. Red is final answer. <a name="a40104z1">
 
log(1+x2)*dx ∫p
=
x*log(1+x2)
x*d[log(1+x2)]
--- eq.LA46
 
=
x*log(1+x2)
x*d(1+x2)

1+x2
=
x*log(1+x2)
x*2x*dx

1+x2
--- eq.LA47
 
=
x*log(1+x2)
2
x2+1-1

1+x2
dx
=
x*log(1+x2)
2
+1+x2

1+x2
dx
2
-1

1+x2
dx
a40104z2
--- eq.LA48
 
=
x*log(1+x2)
2
1*dx
+
2
dx

1+x2
=
x*log(1+x2)
-2x +
2
d[arctan(x)]
--- eq.LA49
 
           
log(1+x2)*dx =    
x*log(1+x2)
-2*x +
2*arctan(x)
+ C
           
--- eq.LA50
width of above equation a401050558
<a name="a40104z3">
2015-01-04-22-16 puzzle at example 6 eq.LA48 . How to get -2x ?
2015-01-05-05-28 review video, at 30:40 find example 6
2015-01-05-05-36 use "+1-1" solve puzzle.
<a name="a40104z4">
2015-01-04-22-27 example 6 equation eq.LA50 
∫log(1+x*x)*dx=x*log(1+x*x)-2x+2*arctan(x)+C 
is in CRC standard mathematical tables 27th ed.
page 277, item 508. 
CRC standard mathematical tables 27th ed.
page 229, item 32 has 
d[arctan(x)]=dx/(1+x*x)
2015-01-04-22-31 

<a name="a40105a">
2015-01-05-08-46 start 
Example 1,2 is xn and sin(x) choose xn as U(x).
Example 3,4 is xn and e3*x , choose xn as U(x).
Example 5,6 above is xn and log(x) , choose log(x) as U(x).
Example 7,8 below is xn and arccos(x),arctan(x) 
choose xn as U(x) ? dxn/dx=n*xn-1
then   arccos(x) antiderivative is x*arccos(x)-√(1-x2)
above CRC 27th p272 item 442, below p229 item 31.
choose arccos(x) as U(x) ? d[arccos(x)]/dx=-1/√(1-x2) 
then   xn antiderivative is xn+1/(n+1)
Consider arccos(x) antiderivative and derivative 
it is better to choose arccos(x) as U(x) !
Because -1/√(1-x2) is easier than x*arccos(x)-√(1-x2)

<a name="a40105b">
Example 7 
Blue is chosen to differentiate. Red is final answer.
 
arccos(x)*dx ∫p
=
x*arccos(x)
x*d[arccos(x)]
--- eq.LA51
 
=
x*arccos(x)
x*
-dx

√(1-x2)
=
x*arccos(x)
+
x*dx

√(1-x2)
--- eq.LA52
 
=
x*arccos(x)
+
(1/2)*d[x2]

√(1-x2)
=
x*arccos(x)
1

2
0 d[x2]

√(1-x2)
a40105b1
--- eq.LA53
 
=
x*arccos(x)
1

2
d[1x2]

√(1-x2)
=
x*arccos(x)
-
1

2
d[√(1-x2) * √(1-x2)]

√(1-x2)
--- eq.LA54
 
=
x*arccos(x)
-
1

2
2*√(1-x2) *d[√(1-x2)]

√(1-x2)
=
x*arccos(x)
-
d[√(1-x2)]
a40105b2
--- eq.LA55
 
               
arccos(x)*dx =    
x*arccos(x)
√(1-x2)
+ C
               
--- eq.LA56
width of above equation a401050942
2015-01-05-09-48
<a name="a40105c">
2015-01-05-12-34 start 
Example 8 below is xn and arctan(x) 
choose xn as U(x) ? dxn/dx=n*xn-1
then   arctan(x) antiderivative is x*arctan(x)-[log(1+x2)]/2
above CRC 27th p272 item 443, below p229 item 32.
choose arctan(x) as U(x) ? d[arctan(x)]/dx=1/(1+x2) 
then   xn antiderivative is xn+1/(n+1)
Consider arctan(x) antiderivative and derivative 
it is better to choose arctan(x) as U(x) !
Because 1/(1+x2) is easier than x*arctan(x)-[log(1+x2)]/2

<a name="a40105d">
Example 8 
Blue is chosen to differentiate. Red is final answer.
 
x*arctan(x)*dx =
arctan(x)*d(x2/2) ∫p
=
x2*arctan(x)

2
x2*d[arctan(x)]

2
--- eq.LA57
 
=
x2*arctan(x)

2
x2*dx

2*(1+x2)
=
x2*arctan(x)

2
x2+1▬1

(1+x2)
dx

2
--- eq.LA58
 
=
x2*arctan(x)

2
1+x2

1+x2
dx

2
▬1

(1+x2)
dx

2
=
x2*arctan(x)

2
dx

2
+
dx

2(1+x2)
a40105d1
--- eq.LA59
 
         
x*arctan(x)*dx
=
x2*arctan(x)

2
x

2
+
arctan(x)

2
+ C          
--- eq.LA60
width of above equation a401051317
2015-01-05-13-18
<a name="a40105e">
2015-01-05-14-10
Above eight examples all involve xn. If given 
two functions u(x) and v(x) not contain xn 
how to determine choosing which one as U(x) ?
Remind: in Integration by parts eq.LA15 
function U(x) first step is differentiation, 
function v(x) first step is integration. 
Next example 9 and 10 the choice of u(x) or 
v(x) does not matter. But need consistent. 
<a name="a40105f">
Example 9 , correct choice ex and ex
Blue is chosen to differentiate. Red is final answer.
 
ex*sin(x)*dx =
ex*d[-cos(x)]
=
ex*d[cos(x)]
--- eq.LA61
 
∫p
=
- [
ex*cos(x)
cos(x)*d[ex]
] =
ex*cos(x)
+
cos(x)*ex*dx
--- eq.LA62
 
=
ex*cos(x)
+
ex*d[sin(x)]
∫p
=
ex*cos(x)
+
ex*sin(x)
sin(x)*d[ex]
a40105f2
--- eq.LA63
 
ex*sin(x)*dx =
ex*cos(x)
+
ex*sin(x)
sin(x)*ex*dx
        two purple ∫
merge to double
--- eq.LA64
 
ex*sin(x)*dx =    
ex[sin(x) - cos(x)]

2
+ C
                               
--- eq.LA65
width of above equation a401051506

<a name="a40105g"> eq.LA66 to eq.LA69 are error on purpose.
Example 9 , wrong choice ex and cos(x)
Above is consistent example 9 , below same example but NOT consistent
Blue is chosen to differentiate. second appearance must choose same.
Below purple color choose another one on purpose, cause trouble.
 
ex*sin(x)*dx =
ex*d[-cos(x)]
=
ex*d[cos(x)]
--- eq.LA66
 
∫p
=
- [
ex*cos(x)
cos(x)*d[ex]
] =
ex*cos(x)
+
cos(x)*ex*dx
--- eq.LA67
 
=
ex*cos(x)
+
cos(x)*d[ex]
∫p
=
ex*cos(x)
+
ex*cos(x)
ex*d[cos(x)]
a40105g2
--- eq.LA68
 
ex*sin(x)*dx =
ex*cos(x)
+
ex*cos(x)
+
sin(x)*ex*dx
        zero=zero
get nothing
--- eq.LA69
width of above equation a401051529
<a name="a40105h">
2015-01-05-15-30 here 
eq.LA66 to eq.LA69 are error on purpose.
In eq.LA66 choose blue ex to differentiate then 
in eq.LA67 choose purple cos(x) to differentiate
this choice is NOT consistent and get zero=zero.
In eq.LA62 choose red ex to differentiate, this 
is correct/consistent choice and get answer 
eq.LA65.
2015-01-05-15-36 

<a name="a40105i">
2015-01-05-15-48 
Above example 9 choose ex to differentiate
Below example 10 choose cos(x) to differentiate

Example 10 , correct choice cos(x) and sin(x)
Blue is chosen to differentiate. Red is final answer.
 
excos(x)*dx =
cos(x)*d[ex]
∫p
=
cos(x)*ex
ex*d[cos(x)]
--- eq.LA70
 
= cos(x)*ex +
exsin(x)*dx
= cos(x)*ex +
sin(x)*d[ex]
--- eq.LA71
 
∫p
=
cos(x)*ex + sin(x)*ex
ex*d[sin(x)]
= cos(x)*ex + sin(x)*ex
excos(x)*dx
a40105i2
--- eq.LA72
 
ex*cos(x)*dx = +
cos(x)*ex
+
sin(x)*ex
excos(x)*dx
        two purple ∫
merge to double
--- eq.LA73
 
ex*cos(x)*dx =    
ex[sin(x) + cos(x)]

2
+ C
                               
--- eq.LA74
width of above equation a401051611

<a name="a40105j"> 2015-01-05-16-12 here Above is ten Integration by parts examples. In Integration by parts eq.LA15 Left side U(x) and v(x) are given functions. Function U(x) first step is differentiation, Function v(x) first step is integration. In two given function choose one as U(x) for differentiation. The priority is 反 arc function arctan(), arcsin() first priority 對 log function second priority 冪 xn function third priority 指 power function ex,5x fourth priority 三 trigonometric function sin(x) last priority Higher priority first step differentiate, Lower priority first step integrate (take antiderivative) nwnu.edu.cn lecturer Professor Hong say the priority in Chinese is 反對冪指三 first appear choose as U(x) record here, in case LiuHH misunderstand. 2015-01-05-16-40 stop

<a name="a40106a"> 2015-01-06-19-52 start Liu,Hsinhan goto google.com search for integration by parts priority found LIATE, ILATE etc. Record is next. 2015-01-06-11-15 http://www.google.com/advanced_search 2015-01-06-11-16 https://www.google.com/search?as_q=priority&as_epq=integration+by+parts+&as_oq=&as_eq=&as_nlo=&as_nhi=&lr=&cr=&as_qdr=all&as_sitesearch=&as_occt=any&safe=images&tbs=&as_filetype=&as_rights=&gws_rd=ssl <a name="a40106b"> 2015-01-06-11-18 https://answers.yahoo.com/question/index?qid=20100801121232AAIIpCM LIATE L logarithms, I inverse trig, arcsin(x) A algebraic, x^3 T trig, sin(x) E exponentials e^x 2015-01-06-11-23 http://www.uea.ac.uk/jtm/12/Lec12p5.pdf LIAE 2015-01-06-11-28 http://www.cuemath.com/iit-jee-mathematics/ii032-introduction-to-integration-by-parts/ ILATE <a name="a40106c"> 2015-01-06-11-31 http://community.boredofstudies.org/14/mathematics-extension-2/112690/integration-parts-help.html LIATE 2015-01-06-11-34 http://math.tutorvista.com/calculus/integration-techniques.html ILATE 2015-01-06-11-43 http://web.science.mq.edu.au/~chris/techniques%20of%20calculus/CHAP04%20Techniques%20of%20Integration.pdf LATE <a name="a40106d"> 2015-01-06-20-03 Integration by parts formula is eq.LA15 U(x) is given and will be differentiated. u(x) is differentiated U(x), u(x) is simpler. v(x) is given and will be integrated. V(x) is integrated v(x), v(x) is not too hard. In ∫f(x)*g(x)*dx we say choose f(x) [OR g(x)] to be U(x), mean we choose one for which will be differentiated. (NOT integrated). Given U(x) must be differentiated to simpler expression and integrated to difficulty expression. Given v(x) is the reverse. v(x) antiderivative V(x) is a little bit harder than v(x).
<a name="a40106e"> Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts
 
x=b
x=a
U(x)v(x)dx =
U(x)V(x)
x=b
x=a
x=b
x=a
V(x)u(x)dx
--- eq.LA15
width of above equation a401062002
<a name="a40106f">
2015-01-06-20-20 here 
To understand the priority of selection of 
f(x) or g(x) as U(x), next is 
Integration by parts priority table
Elementary Functions integrated value and 
differentiated value. Side by side comparison.
<a name="a40106g"> Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts priority table 1
 
highest
priority
as U(x)
if choose
as v(x)
complicated
expression V(x)
or else
choose
as U(x)
  ⇒  
this simpler
formula u(x)
use as v(x)
get more
complicated
f(x)=
arcsin(x)
∫f(x)dx ⇒
  V(x) ⇒
x*arcsin(x)+√(1-x2)
▐ U(x)⇒
df(x)

dx
  ⇒  
1

√(1-x2)
--- eq.LA75
arccos(x)
∫f(x)dx ⇒
  V(x) ⇒
x*arccos(x)-√(1-x2)
▐ U(x)⇒
df(x)

dx
  ⇒  
-1

√(1-x2)
--- eq.LA76
arctan(x)
∫f(x)dx ⇒
  V(x) ⇒
x*arctan(x)
-[log(1+x2)]/2
▐ U(x)⇒
df(x)

dx
  ⇒  
1

1+x2
--- eq.LA77
width of above equation a401062140
2015-01-06-21-40 here
<a name="a40106h">
2015-01-06-22-10 start 
Above is inverse trigonometric function. 
Their integration keep original form and 
pick up additional hard terms. 
If ∫f(x)g(x)dx has inverse trigonometric 
function, choose it as U(x) for first step 
differentiation. 
Next is log function log(x)
<a name="a40106i"> Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts priority table 2
 
highest
priority
as U(x)
if choose
as v(x)
this complicated
expression V(x)
or else
choose
as U(x)
  ⇒  
this simpler
formula u(x)
use as v(x)
get more
complicated
f(x)=
log(x)
∫f(x)dx ⇒
  V(x) ⇒
x*log(x)-x
▐ U(x)⇒
df(x)

dx
  ⇒  
1

x
--- eq.LA78
log(1/x)
∫f(x)dx ⇒
  V(x) ⇒
x*log(1/x)+x
▐ U(x)⇒
df(x)

dx
  ⇒  
-1

x
--- eq.LA79
[log(x)]2
∫f(x)dx ⇒
  V(x) ⇒
x*[log(x)]2+2*x
-2*x*log(x)
▐ U(x)⇒
df(x)

dx
  ⇒  
2*log(x)

x
--- eq.LA80
width of above equation a401062246
2015-01-06-22-46 stop
<a name="a40107a">
2015-01-07-10-28 start 
from 2015-01-07-08-50
 to  2015-01-07-09-50
rewrite Integration by parts priority table 1,2 
id=TLTabl19a eq.LA75 eq.LA76 eq.LA77 
id=TLTabl20a eq.LA78 eq.LA79 eq.LA80 
to 
id=TLTabl21a eq.LA75 eq.LA76 eq.LA77 
id=TLTabl22a eq.LA78 eq.LA79 eq.LA80 
TLTabl19 , TLTabl20 columns not in line.
TLTabl21 , TLTabl22 columns are in line.

<a name="a40107b">
Above is inverse trigonometric function and 
log function log(x) both need differentiate
first, that is both have highest priority 
be chosen as U(x) . Next is 
algebraic function x2 x3
exponential function ex,5x 
trigonometric function sin(x), cos(x)
Entry level reader may confuse 
<a name="a40107c">
What is the difference between algebraic function 
and exponential function ?
In both case 'x' is unknown. 
In both case involve power operation. 
Algebraic function unknown x is base x2
  power 2 is constant.
Exponential function unknown x is power ex
  base e=2.718281828459045 is constant.
Two function differentiation rules and 
integration rules are very different. 
See eq.LA81 and eq.LA82 below. 
2015-01-07-10-48
<a name="a40107d"> Goto Integration by parts derivation begin ; priority table 1,2,3
Integration by parts priority table 3
 
middle
priority
as U(x)
if choose
as v(x)
power increase 1
expression V(x)
or else
choose
as U(x)
  ⇒  
power reduce 1
formula u(x)
use as v(x)
get higher
power
f(x)=
xn
∫f(x)dx ⇒
  V(x) ⇒
xn+1

n+1
▐ U(x)⇒
df(x)

dx
  ⇒  
n*xn-1
--- eq.LA81
change
priority
change
priority
change priority
a40107e
change
priority
change
priority
change
priority
change
priority
change
priority
Lowest
priority
as U(x)
if choose
as v(x)
function V(x)
remain same
or else
choose
as U(x)
  ⇒  
function u(x)
remain same
use as v(x)
not get more
trouble
f(x)=
e(x)
∫f(x)dx ⇒
  V(x) ⇒
e(x)
▐ U(x)⇒
df(x)

dx
  ⇒  
e(x)
--- eq.LA82
sin(x)
∫f(x)dx ⇒
  V(x) ⇒
-cos(x)
▐ U(x)⇒
df(x)

dx
  ⇒  
cos(x)
--- eq.LA83
cos(x)
∫f(x)dx ⇒
  V(x) ⇒
sin(x)
▐ U(x)⇒
df(x)

dx
  ⇒  
-sin(x)
--- eq.LA84
width of above equation a401071115
2015-01-07-11-26 stop
2015-01-07-20-10 done proofread
<a name="a40107f">
2015-01-07-20-12 start 
This file http://freeman2.com/tute0061.htm
is Integration by parts page. This topic 
is an important topic. In the future LiuHH 
may add new material to tute0061.htm 

MrYouMath Gamma Function lecture will be in 
http://freeman2.com/tute0062.htm

<a name="a40107g">
During proofread, in ten example files 

LiuHH added "∫p over =" to distinguish this 
"∫p over =" equality from other equality. 
"∫p over =" mean this step apply 
Integration by parts equation eq.LA15.
2015-01-07-20-22 stop

<a name="a40214a">
2015-02-14-12-52 start 
update 2015-02-14 add links to all Gamma Function
Index topics.
Newer sreies is Taylor Series study notes 
http://freeman2.com/tute0064.htm
2015-02-14-12-55 stop 



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2015-01-05-17-03

Integration by parts ten examples