# Integral #4 (and hopefully a bit more challenging)

#### Random Variable

Challenge Problem:

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0$$

Moderator editor: Approved Challenge question.

#### simplependulum

MHF Hall of Honor
Challenge Problem:

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0$$

Moderator editor: Approved Challenge question.

I am sitting at the library and the time is near to be up , is the answer

$$\displaystyle \Gamma(a+1)\zeta(a+1) \left( 2 - \frac{1}{2^a} \right)$$ ?

#### Drexel28

MHF Hall of Honor
Challenge Problem:

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx, \ a>0$$
$$\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\infty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx$$. But, this is equal to $$\displaystyle 2\int_0^{\infty}e^{-x}x^a\sum_{n=0}^{\infty}e^{-2nx}=2\sum_{n=0}^{\infty}\int_0^\infty x^ae^{-(2n+1)x}$$. But, making the sub $$\displaystyle (2n+1)x=z$$ gives $$\displaystyle 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{a+1}}\int_0^{\infty}z^ae^{-z}dz=2\sum_{n=0}^{\infty}\frac{\Gamma(a+1)}{(2n+1)^{a+1}}=\Gamma(a+1)\zeta(a+1)\left(2-2^{-a}\right)$$

#### Drexel28

MHF Hall of Honor
$$\displaystyle \int_0^{\infty}\frac{x^a}{\sinh(x)}dx=\int_0^{\infty}\frac{x^a}{\frac{e^x-e^{-x}}{2}}=2\int_0^{\infty}\frac{e^{-x}x^a}{1-\left(e^{-x}\right)^2}dx$$. But, this is equal to $$\displaystyle 2\int_0^{\infty}e^{-x}x^a\sum_{n=0}^{\infty}e^{-2nx}=2\sum_{n=0}^{\infty}\int_0^\infty x^ae^{-(2n+1)x}$$. But, making the sub $$\displaystyle (2n+1)x=z$$ gives $$\displaystyle 2\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{a+1}}\int_0^{\infty}z^ae^{-z}dz=2\sum_{n=0}^{\infty}\frac{\Gamma(a+1)}{(2n+1)^{a+1}}=\Gamma(a+1)\zeta(a+1)\left(2-2^{-a}\right)$$
I guess I should probably justify that last sum.

$$\displaystyle \zeta(a+1)=\sum_{n=1}^{\infty}\frac{1}{n^a}=\sum_{n=1}^{\infty}\frac{1}{(2n)^a}+\sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}=\frac{1}{2^a}\zeta(a)+\sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}$$.

Thus, $$\displaystyle \zeta(a)-\frac{1}{2^a}\zeta(a)=\left(1-2^{-a}\right)\zeta(a)=\sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}$$.

So, $$\displaystyle \sum_{n=1}^{\infty}\frac{1}{(2n+1)^a}=\zeta(a+1)\left(1-2^{-(a+1)}\right)=\frac{1}{2}\zeta(a+1)\left(2-2^{-a}\right)$$

#### Random Variable

You guys are too good. I give up. (Tongueout)

Here's what I did:

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^a}{e^{x}-e^{-x}} \ dx = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{e^{2x}-1} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{(e^{x}+1)(e^{x}-1)} \ dx = 2 \int^{\infty}_{0} \frac{(e^{x}+1-1)x^a}{(e^{x}+1)(e^{x}-1)} \ du$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - 2 \int^{\infty}_{0} \frac{x^{a}}{e^{2x}-1} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - \int^{\infty}_{0} \frac{(\frac{u}{2})^{a}}{e^{u}-1} \ dx$$

$$\displaystyle = \Big(2- \frac{1}{2^{a}} \Big) \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx$$

since $$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx = \Gamma(a+1) \zeta(a+1)$$ (an integral which has been done on here before)

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx = \Big(2 - \frac{1}{2^{a}} \Big) \Gamma(a+1) \zeta(a+1)$$

Bruno J.

#### Drexel28

MHF Hall of Honor
You guys are too good. I give up. (Tongueout)

Here's what I did:

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^a}{e^{x}-e^{-x}} \ dx = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{e^{2x}-1} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{e^{x}x^a}{(e^{x}+1)(e^{x}-1)} \ dx = 2 \int^{\infty}_{0} \frac{(e^{x}+1-1)x^a}{(e^{x}+1)(e^{x}-1)} \ du$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - 2 \int^{\infty}_{0} \frac{x^{a}}{e^{2x}-1} \ dx$$

$$\displaystyle = 2 \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx - \int^{\infty}_{0} \frac{(\frac{u}{2})^{a}}{e^{u}-1} \ dx$$

$$\displaystyle = \Big(2- \frac{1}{2^{a}} \Big) \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx$$

since $$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{e^{x}-1} \ dx = \Gamma(a+1) \zeta(a+1)$$ (an integral which has been done on here before)

$$\displaystyle \int^{\infty}_{0} \frac{x^{a}}{\sinh x} \ dx = \Big(2 - \frac{1}{2^{a}} \Big) \Gamma(a+1) \zeta(a+1)$$
Here's one that I have admittedly never tried but looks difficult. How about $$\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n,\text{ }n\in\mathbb{N}$$

#### Bruno J.

MHF Hall of Honor
Here's one that I have admittedly never tried but looks difficult. How about $$\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n,\text{ }n\in\mathbb{N}$$
$$\displaystyle dn$$, right? (Talking)

#### Drexel28

MHF Hall of Honor
$$\displaystyle dn$$, right? (Talking)
Haha, $$\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n{\color{red}dx},\text{ }n\in\mathbb{N}$$

#### NonCommAlg

MHF Hall of Honor
Haha, $$\displaystyle \int_0^{\frac{\pi}{2}}\left(\frac{x}{\sin(x)}\right)^n{\color{red}dx},\text{ }n\in\mathbb{N}$$
can' be done because i couldn't do it! (Rofl) even for n = 1, the value of the integral is in terms of Catalan's constant. so ... i'm just going to leave this integral in peace!

#### simplependulum

MHF Hall of Honor
I think we can obtain a good looking reduction formula separately for even and odd numbers $$\displaystyle n$$ .The key is that can we obtain it for $$\displaystyle n=1,2,3$$ . For $$\displaystyle n =2$$ the integral is $$\displaystyle \pi \ln{2}$$ but for $$\displaystyle n= 1,3$$ the result looks quite bad .... I am looking for a reduction formula for the integral which only consists of two integrals and it only reduces the power of the csc function .