What's the other name for it?

Calculus Level 5

$\large \int_0^\infty \frac{x^5}{e^x+1} \, dx$

If the integral above is equal to $\dfrac AB \pi^C$ for coprime positive integers $A$ and $B$ , find the value of $A+B+C$ .

The answer is 289.

2 solutions

Ayush Verma
Jul 30, 2015

We will use Riemann zeta function and Bernoulli number

$I=\int _{ 0 }^{ \infty }{ \cfrac { { x }^{ 5 } }{ { e }^{ x }+1 } dx= } \int _{ 0 }^{ \infty }{ \cfrac { { e }^{ -x } }{ { 1+e }^{ -x } } { x }^{ 5 }dx } \\ \\ =\int _{ 0 }^{ \infty }{ \left( \sum _{ r=1 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx } } { x }^{ 5 } \right) dx } \\ \\ =\sum _{ r=1 }^{ \infty }{ \left\{ \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } \right\} } \\ \\ let\quad rx=t\Rightarrow \quad \int _{ 0 }^{ \infty }{ { \left( -1 \right) }^{ r-1 }{ e }^{ -rx }{ x }^{ 5 }dx } ={ \left( -1 \right) }^{ r-1 }\int _{ 0 }^{ \infty }{ { e }^{ -t }{ \left( \cfrac { t }{ r } \right) }^{ 6-1 }\cfrac { dt }{ r } } \\ \\ =\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \Gamma 6=120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } \\ \\ \therefore I=\sum _{ r=1 }^{ \infty }{ 120\cfrac { { \left( -1 \right) }^{ r-1 } }{ { r }^{ 6 } } } =120\eta \left( 6 \right) =120\left\{ \left( 1-{ 2 }^{ 1-6 } \right) \right\} \zeta \left( 6 \right) \\ \\ =120\times \cfrac { 31 }{ 32 } \zeta \left( 6 \right) =\cfrac { 465 }{ 4 } \zeta \left( 6 \right) \\ \\ \zeta \left( 6 \right) =\zeta \left( 2\times 3 \right) =\cfrac { { \left( -1 \right) }^{ 3+1 }{ B }_{ 6 }{ \left( 2\pi \right) }^{ 6 } }{ 2\left( 6 \right) ! } =\cfrac { { \pi }^{ 6 } }{ 945 } \quad \quad (as\quad { 6 }^{ th\quad }Bernoulli\quad number{ \quad B }_{ 6 }=\cfrac { 1 }{ 42 } )\\ \\ \therefore I=\cfrac { 465 }{ 4 } \times \cfrac { { \pi }^{ 6 } }{ 945 } =\cfrac { { 31\pi }^{ 6 } }{ 252 } \\ \\ Ans.=A+B+C=31+252+6=289$

Here is a generalisation using the same method.

$I=\int_{0}^{\infty}\frac{x^n}{e^x+1}dx=\int_{0}^{\infty}\frac{e^{-x}x^n}{1+e^{-x}}dx=\int_{0}^{\infty}\sum\limits_{k=1}^{\infty}[(-1)^{k-1}x^ne^{-kx}]dx$

let $kx=u\, \therefore dx=\frac{du}{k}$

$I=\sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{n+1}}\int_{0}^{\infty}u^{n}e^{-u}du=\sum\limits_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^{n+1}}Γ(n+1)\\ \therefore I=\eta(n+1)Γ(n+1)$

Isaac Buckley - 5 years, 10 months ago

Yep. You love generalizing things out and that too some easy thoughts. Nice!

Kartik Sharma - 5 years, 10 months ago

Kartik: do you have an English solution for this?

Pi Han Goh - 5 years, 10 months ago

Shivam Mishra
Mar 23, 2016

well it can be solved pretty easily by using Dirichlet Eta Function

Yup. The other solution poster derived that formula.

Pi Han Goh - 5 years, 2 months ago

What's the other name for it?

The answer is 289.

2 solutions

0 pending reports