Mixing
Integral involving integration by parts
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I have the following integral where the negative exponent on natural e is making me question my approach.
Problem:
begineqnarray*
int_0^infty x^3e^-x dx.
endeqnarray*
My approach so far:
- I am assuming I am to use $u$-sub or integration by parts.
- I have $u = -x$
- I end up getting to $u^3e^u-int3u^2e^u,du$
I am stuck here. Any advice would be greatly appreciated. Also, a confirmation of my approach would be helpful. Am I missing any technique due to the negative exponent?
calculus integration definite-integrals
edited Aug 2 at 22:55
Robert Howard
1,263620
asked Aug 2 at 20:19
jackbenimbo
476
 |Â
show 3 more comments
up vote
0
down vote
favorite
I have the following integral where the negative exponent on natural e is making me question my approach.
Problem:
begineqnarray*
int_0^infty x^3e^-x dx.
endeqnarray*
My approach so far:
- I am assuming I am to use $u$-sub or integration by parts.
- I have $u = -x$
- I end up getting to $u^3e^u-int3u^2e^u,du$
I am stuck here. Any advice would be greatly appreciated. Also, a confirmation of my approach would be helpful. Am I missing any technique due to the negative exponent?
calculus integration definite-integrals
edited Aug 2 at 22:55
Robert Howard
1,263620
asked Aug 2 at 20:19
jackbenimbo
476
2
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
2
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
1
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
2
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
1
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have the following integral where the negative exponent on natural e is making me question my approach.
Problem:
begineqnarray*
int_0^infty x^3e^-x dx.
endeqnarray*
My approach so far:
- I am assuming I am to use $u$-sub or integration by parts.
- I have $u = -x$
- I end up getting to $u^3e^u-int3u^2e^u,du$
I am stuck here. Any advice would be greatly appreciated. Also, a confirmation of my approach would be helpful. Am I missing any technique due to the negative exponent?
calculus integration definite-integrals
edited Aug 2 at 22:55
Robert Howard
1,263620
asked Aug 2 at 20:19
jackbenimbo
476
I have the following integral where the negative exponent on natural e is making me question my approach.
Problem:
begineqnarray*
int_0^infty x^3e^-x dx.
endeqnarray*
My approach so far:
- I am assuming I am to use $u$-sub or integration by parts.
- I have $u = -x$
- I end up getting to $u^3e^u-int3u^2e^u,du$
I am stuck here. Any advice would be greatly appreciated. Also, a confirmation of my approach would be helpful. Am I missing any technique due to the negative exponent?
calculus integration definite-integrals
edited Aug 2 at 22:55
Robert Howard
1,263620
asked Aug 2 at 20:19
jackbenimbo
476
edited Aug 2 at 22:55
Robert Howard
1,263620
edited Aug 2 at 22:55
Robert Howard
1,263620
edited Aug 2 at 22:55
Robert Howard
1,263620
1,263620
asked Aug 2 at 20:19
jackbenimbo
476
asked Aug 2 at 20:19
jackbenimbo
476
asked Aug 2 at 20:19
jackbenimbo
476
476
2
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
2
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
1
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
2
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
1
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19
 |Â
show 3 more comments
2
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
2
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
1
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
2
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
1
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19
2
2
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
2
2
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
1
1
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
2
2
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
1
1
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19
 |Â
show 3 more comments
5 Answers
5
active
oldest
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up vote
0
down vote
accepted
$$int x^3e^-xdx=-x^3e^-x+int3x^2e^-xdx=-x^3e^-x-3x^2e^-x+6int xe^-xdx=$$
$$=-x^3e^-x-3x^2e^-x-6xe^-x+6int e^-xdx=-x^3e^-x-3x^2e^-x-6xe^-x-6e^-x+C.$$
Can you end it now?
I got the answer: $6$.
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
add a comment |Â
up vote
2
down vote
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = int_0^infty x^n e^-x dx$ then by integration by parts show that $I_n+1 = (n+1)I_n,$ therefore $I_n = n!int_0^infty e^-x dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = int_0^infty e^-sx dx = frac 1s$
$frac dds f(s) = int_0^infty frac dds e^-sx dx = int_0^infty -xe^-sx dx = frac dds frac 1s = - s^-2\
frac d^nds^n f(s) = int_0^infty (-1)^nx^ne^-sx dx = (-1)^n n! s^-(n+1)$
Evaluate at $s= 1$
$int_0^infty x^ne^-x dx = n!$
And eventually you will learn about the $Gamma$ function and the Laplace Transform...
answered Aug 2 at 20:59
Doug M
39k31748
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
add a comment |Â
up vote
1
down vote
Using the substitution $x=-log y$ the integral $I= int_0^infty x^3 ; e^-x , dx$ transforms to
$$I=-int_0^1 ; (log y)^3 ; dy$$
However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(log x)^3$ in terms of the integral of $(log x)^2$
$$int (log x)^3 ,dx=log x int (log x)^2 ,dx,-int frac1x , int (log x)^2 , dx ,dxtag1$$
The integral of $(log x)^2$ in terms of the integral of $log x$
$$int (log x)^2 ,dx=log x int log x ,dx,-int frac1x , int log x , dx ,dxtag2$$
The integral of $log x$ in terms of the integral of $1$
$$int log x ,dx=log xint,1,dx-int frac1xint,1,dx,dx=x log x -xtag3$$
The full calculation is made in this question Integral of $(log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
answered Aug 2 at 23:01
James Arathoon
1,155420
add a comment |Â
up vote
1
down vote
One more method (when you don't want to do multiple IBP):
$$int x^3e^-x dx=C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5 iff \
(C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5)'=x^3e^-x iff \
colorgreen3C_1x^2e^-x-colorredC_1x^3e^-x+colorblue2C_2xe^-x-colorgreenC_2x^2e^-x+colorbrownC_3e^-x-colorblueC_3xe^-x-colorbrownC_4e^-x=x^3e^-x iff \
begincasescolorred-C_1=1\
colorgreen3C_1-C_2=0\
colorblue2C_2-C_3=0\
colorbrownC_3-C_4=0endcases Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\
$$
answered Aug 3 at 15:05
farruhota
13.5k2632
add a comment |Â
up vote
1
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$$I=int_0^inftyx^3e^-xdx$$
There are two main ways we can approach this, firstly is the gamma function. It is known that:
$$n!=Gamma(n+1)=int_0^inftyx^ne^-xdx$$
$$therefore,I=Gamma(4)=3!=6$$
The second way is indeed integration by parts. To recap:
$$int ufracdvdxdx=uv-int vfracdudxdx$$
Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion:
$$I=int_0^inftyx^3e^-xdx$$
$w=-x,,dx=-dw$
$$therefore,I=int_0^-inftyw^3e^wdw=left[w^3e^wright]_0^-infty-3int_0^-inftyw^2e^wdw=-3left(left[w^2e^wright]_0^-infty-2int_0^-inftywe^wdwright)=6int_0^-inftywe^wdw=6left(left[we^wright]_0^-infty-int_0^-inftye^wdwright)=-6int_0^-inftye^wdw=6int_-infty^0e^wdw=6left[e^wright]_-infty^0=6(1-0)=6$$
answered 21 hours ago
Henry Lee
49210
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
$$int x^3e^-xdx=-x^3e^-x+int3x^2e^-xdx=-x^3e^-x-3x^2e^-x+6int xe^-xdx=$$
$$=-x^3e^-x-3x^2e^-x-6xe^-x+6int e^-xdx=-x^3e^-x-3x^2e^-x-6xe^-x-6e^-x+C.$$
Can you end it now?
I got the answer: $6$.
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
add a comment |Â
up vote
0
down vote
accepted
$$int x^3e^-xdx=-x^3e^-x+int3x^2e^-xdx=-x^3e^-x-3x^2e^-x+6int xe^-xdx=$$
$$=-x^3e^-x-3x^2e^-x-6xe^-x+6int e^-xdx=-x^3e^-x-3x^2e^-x-6xe^-x-6e^-x+C.$$
Can you end it now?
I got the answer: $6$.
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
$$int x^3e^-xdx=-x^3e^-x+int3x^2e^-xdx=-x^3e^-x-3x^2e^-x+6int xe^-xdx=$$
$$=-x^3e^-x-3x^2e^-x-6xe^-x+6int e^-xdx=-x^3e^-x-3x^2e^-x-6xe^-x-6e^-x+C.$$
Can you end it now?
I got the answer: $6$.
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
$$int x^3e^-xdx=-x^3e^-x+int3x^2e^-xdx=-x^3e^-x-3x^2e^-x+6int xe^-xdx=$$
$$=-x^3e^-x-3x^2e^-x-6xe^-x+6int e^-xdx=-x^3e^-x-3x^2e^-x-6xe^-x-6e^-x+C.$$
Can you end it now?
I got the answer: $6$.
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
answered Aug 2 at 20:59
Michael Rozenberg
87.2k1577178
87.2k1577178
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
add a comment |Â
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
Thanks for the step by step illustration.
– jackbenimbo
Aug 2 at 21:14
1
1
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
You are welcome! The integration by parts forever!
– Michael Rozenberg
Aug 2 at 21:15
add a comment |Â
up vote
2
down vote
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = int_0^infty x^n e^-x dx$ then by integration by parts show that $I_n+1 = (n+1)I_n,$ therefore $I_n = n!int_0^infty e^-x dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = int_0^infty e^-sx dx = frac 1s$
$frac dds f(s) = int_0^infty frac dds e^-sx dx = int_0^infty -xe^-sx dx = frac dds frac 1s = - s^-2\
frac d^nds^n f(s) = int_0^infty (-1)^nx^ne^-sx dx = (-1)^n n! s^-(n+1)$
Evaluate at $s= 1$
$int_0^infty x^ne^-x dx = n!$
And eventually you will learn about the $Gamma$ function and the Laplace Transform...
answered Aug 2 at 20:59
Doug M
39k31748
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
add a comment |Â
up vote
2
down vote
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = int_0^infty x^n e^-x dx$ then by integration by parts show that $I_n+1 = (n+1)I_n,$ therefore $I_n = n!int_0^infty e^-x dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = int_0^infty e^-sx dx = frac 1s$
$frac dds f(s) = int_0^infty frac dds e^-sx dx = int_0^infty -xe^-sx dx = frac dds frac 1s = - s^-2\
frac d^nds^n f(s) = int_0^infty (-1)^nx^ne^-sx dx = (-1)^n n! s^-(n+1)$
Evaluate at $s= 1$
$int_0^infty x^ne^-x dx = n!$
And eventually you will learn about the $Gamma$ function and the Laplace Transform...
answered Aug 2 at 20:59
Doug M
39k31748
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = int_0^infty x^n e^-x dx$ then by integration by parts show that $I_n+1 = (n+1)I_n,$ therefore $I_n = n!int_0^infty e^-x dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = int_0^infty e^-sx dx = frac 1s$
$frac dds f(s) = int_0^infty frac dds e^-sx dx = int_0^infty -xe^-sx dx = frac dds frac 1s = - s^-2\
frac d^nds^n f(s) = int_0^infty (-1)^nx^ne^-sx dx = (-1)^n n! s^-(n+1)$
Evaluate at $s= 1$
$int_0^infty x^ne^-x dx = n!$
And eventually you will learn about the $Gamma$ function and the Laplace Transform...
answered Aug 2 at 20:59
Doug M
39k31748
The direct way to do this is to integrate by parts 3 times. Each time reducing the exponent of $x$ by $1$
A little sneaker is to say let $I_n = int_0^infty x^n e^-x dx$ then by integration by parts show that $I_n+1 = (n+1)I_n,$ therefore $I_n = n!int_0^infty e^-x dx = n!.$
If you hate integration by parts, you can use differentiation under the integral sign, aka "Feynman's trick"
let $f(s) = int_0^infty e^-sx dx = frac 1s$
$frac dds f(s) = int_0^infty frac dds e^-sx dx = int_0^infty -xe^-sx dx = frac dds frac 1s = - s^-2\
frac d^nds^n f(s) = int_0^infty (-1)^nx^ne^-sx dx = (-1)^n n! s^-(n+1)$
Evaluate at $s= 1$
$int_0^infty x^ne^-x dx = n!$
And eventually you will learn about the $Gamma$ function and the Laplace Transform...
answered Aug 2 at 20:59
Doug M
39k31748
answered Aug 2 at 20:59
Doug M
39k31748
answered Aug 2 at 20:59
Doug M
39k31748
answered Aug 2 at 20:59
Doug M
39k31748
39k31748
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
add a comment |Â
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
Thank you for your contribution! Is this something we learn in Calc II (Early Transcendantals - briggs text) or perhaps later in Vector Calc..?
– jackbenimbo
Aug 2 at 21:13
1
1
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
It really depends on your instructor and what they think is important. There are many "integration tricks" which do not require any more theoretical underpinning than you already have, but which an instruct may not think is worth the time do discuss.
– Doug M
Aug 2 at 21:43
add a comment |Â
up vote
1
down vote
Using the substitution $x=-log y$ the integral $I= int_0^infty x^3 ; e^-x , dx$ transforms to
$$I=-int_0^1 ; (log y)^3 ; dy$$
However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(log x)^3$ in terms of the integral of $(log x)^2$
$$int (log x)^3 ,dx=log x int (log x)^2 ,dx,-int frac1x , int (log x)^2 , dx ,dxtag1$$
The integral of $(log x)^2$ in terms of the integral of $log x$
$$int (log x)^2 ,dx=log x int log x ,dx,-int frac1x , int log x , dx ,dxtag2$$
The integral of $log x$ in terms of the integral of $1$
$$int log x ,dx=log xint,1,dx-int frac1xint,1,dx,dx=x log x -xtag3$$
The full calculation is made in this question Integral of $(log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
answered Aug 2 at 23:01
James Arathoon
1,155420
add a comment |Â
up vote
1
down vote
Using the substitution $x=-log y$ the integral $I= int_0^infty x^3 ; e^-x , dx$ transforms to
$$I=-int_0^1 ; (log y)^3 ; dy$$
However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(log x)^3$ in terms of the integral of $(log x)^2$
$$int (log x)^3 ,dx=log x int (log x)^2 ,dx,-int frac1x , int (log x)^2 , dx ,dxtag1$$
The integral of $(log x)^2$ in terms of the integral of $log x$
$$int (log x)^2 ,dx=log x int log x ,dx,-int frac1x , int log x , dx ,dxtag2$$
The integral of $log x$ in terms of the integral of $1$
$$int log x ,dx=log xint,1,dx-int frac1xint,1,dx,dx=x log x -xtag3$$
The full calculation is made in this question Integral of $(log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
answered Aug 2 at 23:01
James Arathoon
1,155420
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Using the substitution $x=-log y$ the integral $I= int_0^infty x^3 ; e^-x , dx$ transforms to
$$I=-int_0^1 ; (log y)^3 ; dy$$
However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(log x)^3$ in terms of the integral of $(log x)^2$
$$int (log x)^3 ,dx=log x int (log x)^2 ,dx,-int frac1x , int (log x)^2 , dx ,dxtag1$$
The integral of $(log x)^2$ in terms of the integral of $log x$
$$int (log x)^2 ,dx=log x int log x ,dx,-int frac1x , int log x , dx ,dxtag2$$
The integral of $log x$ in terms of the integral of $1$
$$int log x ,dx=log xint,1,dx-int frac1xint,1,dx,dx=x log x -xtag3$$
The full calculation is made in this question Integral of $(log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
answered Aug 2 at 23:01
James Arathoon
1,155420
Using the substitution $x=-log y$ the integral $I= int_0^infty x^3 ; e^-x , dx$ transforms to
$$I=-int_0^1 ; (log y)^3 ; dy$$
However the most obvious route to evaluating this integral still requires integration by parts 3 times plus a lot of additional algebra
The integral of $(log x)^3$ in terms of the integral of $(log x)^2$
$$int (log x)^3 ,dx=log x int (log x)^2 ,dx,-int frac1x , int (log x)^2 , dx ,dxtag1$$
The integral of $(log x)^2$ in terms of the integral of $log x$
$$int (log x)^2 ,dx=log x int log x ,dx,-int frac1x , int log x , dx ,dxtag2$$
The integral of $log x$ in terms of the integral of $1$
$$int log x ,dx=log xint,1,dx-int frac1xint,1,dx,dx=x log x -xtag3$$
The full calculation is made in this question Integral of $(log x)^3$ (Spivak's Calculus, Chapter 19, Problem 3v) plus answers along the lines already investigated.
answered Aug 2 at 23:01
James Arathoon
1,155420
answered Aug 2 at 23:01
James Arathoon
1,155420
answered Aug 2 at 23:01
James Arathoon
1,155420
answered Aug 2 at 23:01
James Arathoon
1,155420
1,155420
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One more method (when you don't want to do multiple IBP):
$$int x^3e^-x dx=C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5 iff \
(C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5)'=x^3e^-x iff \
colorgreen3C_1x^2e^-x-colorredC_1x^3e^-x+colorblue2C_2xe^-x-colorgreenC_2x^2e^-x+colorbrownC_3e^-x-colorblueC_3xe^-x-colorbrownC_4e^-x=x^3e^-x iff \
begincasescolorred-C_1=1\
colorgreen3C_1-C_2=0\
colorblue2C_2-C_3=0\
colorbrownC_3-C_4=0endcases Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\
$$
answered Aug 3 at 15:05
farruhota
13.5k2632
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up vote
1
down vote
One more method (when you don't want to do multiple IBP):
$$int x^3e^-x dx=C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5 iff \
(C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5)'=x^3e^-x iff \
colorgreen3C_1x^2e^-x-colorredC_1x^3e^-x+colorblue2C_2xe^-x-colorgreenC_2x^2e^-x+colorbrownC_3e^-x-colorblueC_3xe^-x-colorbrownC_4e^-x=x^3e^-x iff \
begincasescolorred-C_1=1\
colorgreen3C_1-C_2=0\
colorblue2C_2-C_3=0\
colorbrownC_3-C_4=0endcases Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\
$$
answered Aug 3 at 15:05
farruhota
13.5k2632
add a comment |Â
up vote
1
down vote
up vote
1
down vote
One more method (when you don't want to do multiple IBP):
$$int x^3e^-x dx=C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5 iff \
(C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5)'=x^3e^-x iff \
colorgreen3C_1x^2e^-x-colorredC_1x^3e^-x+colorblue2C_2xe^-x-colorgreenC_2x^2e^-x+colorbrownC_3e^-x-colorblueC_3xe^-x-colorbrownC_4e^-x=x^3e^-x iff \
begincasescolorred-C_1=1\
colorgreen3C_1-C_2=0\
colorblue2C_2-C_3=0\
colorbrownC_3-C_4=0endcases Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\
$$
answered Aug 3 at 15:05
farruhota
13.5k2632
One more method (when you don't want to do multiple IBP):
$$int x^3e^-x dx=C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5 iff \
(C_1x^3e^-x+C_2x^2e^-x+C_3xe^-x+C_4e^-x+C_5)'=x^3e^-x iff \
colorgreen3C_1x^2e^-x-colorredC_1x^3e^-x+colorblue2C_2xe^-x-colorgreenC_2x^2e^-x+colorbrownC_3e^-x-colorblueC_3xe^-x-colorbrownC_4e^-x=x^3e^-x iff \
begincasescolorred-C_1=1\
colorgreen3C_1-C_2=0\
colorblue2C_2-C_3=0\
colorbrownC_3-C_4=0endcases Rightarrow C_1=-1,C_2=-3,C_3=-6, C_4=-6.\
$$
answered Aug 3 at 15:05
farruhota
13.5k2632
answered Aug 3 at 15:05
farruhota
13.5k2632
answered Aug 3 at 15:05
farruhota
13.5k2632
answered Aug 3 at 15:05
farruhota
13.5k2632
13.5k2632
add a comment |Â
add a comment |Â
up vote
1
down vote
$$I=int_0^inftyx^3e^-xdx$$
There are two main ways we can approach this, firstly is the gamma function. It is known that:
$$n!=Gamma(n+1)=int_0^inftyx^ne^-xdx$$
$$therefore,I=Gamma(4)=3!=6$$
The second way is indeed integration by parts. To recap:
$$int ufracdvdxdx=uv-int vfracdudxdx$$
Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion:
$$I=int_0^inftyx^3e^-xdx$$
$w=-x,,dx=-dw$
$$therefore,I=int_0^-inftyw^3e^wdw=left[w^3e^wright]_0^-infty-3int_0^-inftyw^2e^wdw=-3left(left[w^2e^wright]_0^-infty-2int_0^-inftywe^wdwright)=6int_0^-inftywe^wdw=6left(left[we^wright]_0^-infty-int_0^-inftye^wdwright)=-6int_0^-inftye^wdw=6int_-infty^0e^wdw=6left[e^wright]_-infty^0=6(1-0)=6$$
answered 21 hours ago
Henry Lee
49210
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
add a comment |Â
up vote
1
down vote
$$I=int_0^inftyx^3e^-xdx$$
There are two main ways we can approach this, firstly is the gamma function. It is known that:
$$n!=Gamma(n+1)=int_0^inftyx^ne^-xdx$$
$$therefore,I=Gamma(4)=3!=6$$
The second way is indeed integration by parts. To recap:
$$int ufracdvdxdx=uv-int vfracdudxdx$$
Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion:
$$I=int_0^inftyx^3e^-xdx$$
$w=-x,,dx=-dw$
$$therefore,I=int_0^-inftyw^3e^wdw=left[w^3e^wright]_0^-infty-3int_0^-inftyw^2e^wdw=-3left(left[w^2e^wright]_0^-infty-2int_0^-inftywe^wdwright)=6int_0^-inftywe^wdw=6left(left[we^wright]_0^-infty-int_0^-inftye^wdwright)=-6int_0^-inftye^wdw=6int_-infty^0e^wdw=6left[e^wright]_-infty^0=6(1-0)=6$$
answered 21 hours ago
Henry Lee
49210
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$$I=int_0^inftyx^3e^-xdx$$
There are two main ways we can approach this, firstly is the gamma function. It is known that:
$$n!=Gamma(n+1)=int_0^inftyx^ne^-xdx$$
$$therefore,I=Gamma(4)=3!=6$$
The second way is indeed integration by parts. To recap:
$$int ufracdvdxdx=uv-int vfracdudxdx$$
Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion:
$$I=int_0^inftyx^3e^-xdx$$
$w=-x,,dx=-dw$
$$therefore,I=int_0^-inftyw^3e^wdw=left[w^3e^wright]_0^-infty-3int_0^-inftyw^2e^wdw=-3left(left[w^2e^wright]_0^-infty-2int_0^-inftywe^wdwright)=6int_0^-inftywe^wdw=6left(left[we^wright]_0^-infty-int_0^-inftye^wdwright)=-6int_0^-inftye^wdw=6int_-infty^0e^wdw=6left[e^wright]_-infty^0=6(1-0)=6$$
answered 21 hours ago
Henry Lee
49210
$$I=int_0^inftyx^3e^-xdx$$
There are two main ways we can approach this, firstly is the gamma function. It is known that:
$$n!=Gamma(n+1)=int_0^inftyx^ne^-xdx$$
$$therefore,I=Gamma(4)=3!=6$$
The second way is indeed integration by parts. To recap:
$$int ufracdvdxdx=uv-int vfracdudxdx$$
Firstly for our integral it may be easier to make a substitution to remove the minus sign to avoid confusion:
$$I=int_0^inftyx^3e^-xdx$$
$w=-x,,dx=-dw$
$$therefore,I=int_0^-inftyw^3e^wdw=left[w^3e^wright]_0^-infty-3int_0^-inftyw^2e^wdw=-3left(left[w^2e^wright]_0^-infty-2int_0^-inftywe^wdwright)=6int_0^-inftywe^wdw=6left(left[we^wright]_0^-infty-int_0^-inftye^wdwright)=-6int_0^-inftye^wdw=6int_-infty^0e^wdw=6left[e^wright]_-infty^0=6(1-0)=6$$
answered 21 hours ago
Henry Lee
49210
answered 21 hours ago
Henry Lee
49210
answered 21 hours ago
Henry Lee
49210
answered 21 hours ago
Henry Lee
49210
49210
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
add a comment |Â
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
1
1
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
Thanks you for your clear approach, Henry. I appreciate you taking the time.
– jackbenimbo
15 hours ago
add a comment |Â
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2
You approach will work. You just have to do integration by parts two more times. I think most people wouldn't have bothered with the $u$-substitution.
– B. Goddard
Aug 2 at 20:21
2
Keep integrating by parts until the $x^3$ differentiates to $6$.
– Donald Splutterwit
Aug 2 at 20:27
1
@jackbenimbo Yes, that's correct.
– John
Aug 2 at 20:37
2
I would use the by-parts shortcut of tabular integration (also known as the "Stand and Deliver" method, because it showed up in that movie.
– Adrian Keister
Aug 2 at 20:51
1
Thank you Robert! I had no idea this was available. This way I can make them more clear in the future, no doubt. Appreciate it.
– jackbenimbo
Aug 2 at 21:19