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

```<a name="Gamma_Indx"> t061link()
2015-01-01-15-18 start
On
2014-03-25-20-36 Liu,Hsinhan access

Gamma Function
•by MrYouMath
•12 videos
•5,940 views
•1 hour, 48 minutes
Gamma Function - Part 1 - Functional Equation
Gamma Function - Part 2 - Gauss Representation
Gamma Function - Part 3 - Weierstrass Representation
Gamma Function - Part 4 - Relationship to Sine
Gamma Function - Part 5 - Gamma of 0.5 ( one half)
Gamma Function - Part 6 - Stirling's Approximation
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
Gamma Function - Part 8 - Euler Integral II The Sinc-Function
Gamma Function - Part 9 - Euler Integral III Fresnel Integral
Gamma Function - Part 10 - Beta Function
Gamma Function - Part 11 - Legendre Duplication Formula
Gamma Function - Part 12 - Relation to Zeta Function
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

2014-10-29-07-56
2014-10-29-08-24
MrYouMath_17_files_Riemann.htm
"Zeta Function - Part 1 - Convergence"
"Zeta Function - Part 2 - Euler Product Representation"
"Zeta Function - Part 3 - Euler Product (easy)"
<a name="docA007">
"Zeta Function - Part 4 - Infinitude of Prime Numbers"
"Zeta Function - Part 5 - Prime Zeta Function"
"Zeta Function - Part 6 - The Prime Counting Function"
"Zeta Function - Part 7 - Zeta of 2 aka The Basel Problem"
"Zeta Function - Part 8 - Zeta of 2n - Part 1"
<a name="docA008">
"Zeta Function - Part 8 - Zeta of 2n - Part 2"
"Zeta Function - Part 8 - Zeta of 2n - Part 3"
"Zeta Function - Part 9 - Relation to Gamma Function"
"Zeta Function - Part 10 - Jacobi Theta Function"
"Zeta Function - Part 11 - Riemann Functional Equation I"
<a name="docA009">
"Zeta Function - Part 12 - Riemann Functional Equation II"
"Zeta Function - Part 13 - Trivial Zeros of the Zeta Function"
"Zeta Function - Part 14 - Riemann Xi Function"
"Sine Function Product Formula (Hadamard Factorization Theorem)"
"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,
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 ?
```
<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.
exam each word.
Below is mathematics teacher's lecture.
This is a Chinese video. URL valid in 2013.
<a name="a40104s">examp10()

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">examp10()
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">examp10()
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">examp10()
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">examp10()
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">examp10()
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">examp10()
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">examp10()
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[1▬x2] √(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">examp10()
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">examp10()
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

Higher priority first step differentiate,
Lower priority first step integrate (take antiderivative)

nwnu.edu.cn lecturer Professor Hong say
the priority in Chinese is

record here, in case LiuHH misunderstand.
2015-01-05-16-40 stop

<a name="a40106a">
2015-01-06-19-52 start
integration by parts
priority
found LIATE, ILATE etc. Record is next.

2015-01-06-11-15

2015-01-06-11-16

<a name="a40106b">
2015-01-06-11-18
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
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)
<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
```<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
examp10()
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
Index topics.
Newer sreies is Taylor Series study notes
http://freeman2.com/tute0064.htm
2015-02-14-12-55 stop

Following is frequently needed strings.
f'(x)=limit[h→0]{[C-C]/[(x+h)-x]}
xn x2 πz ●●●
π Γ ∫ ζ(s) e-t ξ() zm
Riemann Functional Equation
y2+2i*y*k/x  ∏[n=1,∞]  i=√(-1) √

```

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file name tute0061.htm mean
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2015-01-01-20-45 save as tute0061.htm

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http://freeman2.com/tute0061.htm