Evaluating the following integral: $\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$

Question

How can we compute this integral for all $\operatorname{Re}(a)>0$ and $\operatorname{Re}(b)>0$?

$$\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx$$

Is there a way to compute it using methods from real analysis?

Answer 1

We first start by converting our integral into a sum. We have that (by the geometric series): $$\frac{1}{e^{bx}-1}=\sum_{n=1}^{\infty} e^{-bxn}$$ Additionally, we have that (as can be found by simply finding an antiderivative): $$\int_0^{\infty} x\cos(ax)e^{-bxn}~dx=\frac{-a^2+b^2 n^2}{(a^2+b^2 n^2)^2}$$ Therefore, the integral in question is equal to the following sum: $$\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx=\sum_{n=1}^{\infty} \frac{-a^2+b^2n^2}{(a^2+b^2 n^2)^2}=-a^2 S_1+b^2 S_2 \tag{1}$$ Therefore, it suffices to compute the following sums: $$S_1:=\sum_{n=1}^{\infty} \frac{1}{(a^2+b^2 n^2)^2},\quad S_2:=\sum_{n=1}^{\infty} \frac{n^2}{(a^2+b^2 n^2)^2}$$ To do this, we start from the well-known result that (which can be derived using real analysis, as shown by several answers): $$\sum_{n=0}^\infty\frac{1}{c^2+n^2}=\frac{1+c\pi\coth (c\pi)}{2c^2}$$ From this, by using the substitution $c:=a/b$ one can easily derive the more general result that (note the difference in the lower limit of the sum): $$\sum_{n=1}^{\infty} \frac{1}{a^2+b^2 n^2}=\frac{-b+a\pi \coth(a\pi/b)}{2a^2 b} \tag{2}$$ Now, we can compute closed forms for the sums $S_1$ and $S_2$ by differentiating $(2)$ with respect to $a$ and $b$ respectively. We hence have that: $$S_1=\frac{\pi^2 a^2 \operatorname{csch}^2(a\pi /b)+\pi a b \coth(a\pi /b)-2b^2}{4a^4 b^2}$$ $$S_2=\frac{-\pi^2 a^2 \operatorname{csch}^2(a\pi /b)+\pi a b \coth(a\pi /b)}{4a^2 b^4}$$ Therefore, by $(1)$, the integral is given by: $$\bbox[5px,border:2px solid #C0A000]{\int_{0}^{\infty }\frac{x\cos(ax)}{e^{bx}-1}\ dx=\frac{1}{2} \left(\frac{1}{a^2}-\frac{\pi ^2 \operatorname{csch}^2(a\pi/b)}{b^2}\right)}$$

Answer 2

Let $$ I(a)=\int_{0}^{\infty}\frac{\sin(ax)}{e^{bx}-1}\ dx. $$ It is easy to see $$ I'(a)=\int_{0}^{\infty}\frac{x\cos(ax)}{e^{bx}-1}\ dx. $$ Since \begin{eqnarray} I(a)&=&\int_{0}^{\infty}\frac{e^{-bx}\sin(ax)}{1-e^{-bx}}\ dx=\int_{0}^{\infty}\sum_{n=0}^\infty e^{-b(n+1)x}\sin(ax)\ dx\\ &=&\sum_{n=0}^\infty\int_{0}^{\infty} e^{-b(n+1)x}\sin(ax)\ dx=\sum_{n=0}^\infty\frac{a}{a^2+b^2(n+1)^2}. \end{eqnarray} Using the result from @projectilemotion, one has \begin{eqnarray} I(a)&=&\sum_{n=0}^\infty\frac{a}{a^2+b^2(n+1)^2}=\frac{-b+a\pi \coth(a\pi/b)}{2a b}=\frac12\left(-\frac{1}{a}+\frac\pi b\coth(\frac{a\pi}{b})\right) \end{eqnarray} and hence $$ I'(a)=\frac12\left(\frac1{a^2}-\frac{\pi^2}{b^2}\text{csch}^2(\frac{a\pi}{b})\right).$$

Answer 3

$$\displaystyle{\displaylines{I=\int_{0}^{\infty }\frac{x cos(ax)}{e^{bx} -1} dx =\int_{0}^{\infty }\frac{xe^{-bx}cos(ax)}{1-e^{-bx}} dx , \forall b\gt 0}}$$ It's easy to see : $$\displaystyle{\displaylines{I=\sum_{n=1}^{\infty }\int_{0}^{\infty } xe^{-bnx} . cos(ax)dx=\frac{\partial }{\partial a}\sum_{n=1}^{\infty }\int_{0}^{\infty } e^{-bnx} .sin(ax)dx}}$$ so we have ; $$\displaystyle{\displaylines{I=\frac{\partial }{\partial a}\sum_{n=1}^{\infty }\frac{a}{a^{2}+b^{2}.n^{2}}}}$$ But :

$$\displaystyle{\displaylines{\sum_{n=1}^{\infty }\frac{1}{x^{2}+n^{2}}=\frac{-1}{2x^{2}}+\frac{\pi(e^{x\pi}+e^{-x\pi})}{2x(e^{x\pi}-e^{-x\pi})}}}$$ we can write : $$\displaystyle{\displaylines{\sum_{n=1}^{\infty }\frac{a}{a^{2}+b^{2}.n^{2}}=\frac{a}{b^{2}}\sum_{n=1}^{\infty }\frac{1}{(\frac{a}{b})^{2}+n^{2}}=\frac{1}{a}\left[ (\frac{a}{b})^{2}\sum_{n=1}^{\infty }\frac{1}{(\frac{a}{b})^{2}+n^{2}} \right]}}$$

$$=\frac{1}{a}\left[ \frac{a^{2}}{b^{2}} \left( \frac{-1}{2(\frac{a}{b})^{2}} +\frac{\pi(e^{\frac{a}{b}\pi}+e^{\frac{-a}{b}\pi})}{2(\frac{a}{b})(e^{\frac{a}{b}\pi}-e^{\frac{-a}{b}\pi})}\right)\right]$$

so we have :$$\displaystyle{\displaylines{\sum_{n=1}^{\infty }\frac{a}{a^{2}+b^{2}.n^{2}}=\frac{-1}{2a}+\frac{\pi}{2b}.coth(\frac{a\pi}{b})}}$$ By partial derivative for both sides :

$$\displaystyle{\displaylines{\int_{0}^{\infty }\frac{xcos(ax)}{e^{bx} -1} dx =\frac{1}{2a^{2}}-\frac{\pi^{2}}{2b^{2}} csch^{2}(\frac{a\pi}{b})}}$$

Answer 4

**my attempt **

$$I=\int_{0}^{\infty }\frac{x\ cos(ax)}{e^{bx}-1}dx=\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}cos(ax)dx\\ \\ \\ =\frac{1}{2}\sum_{n=1}^{\infty }\int_{0}^{\infty }x\ e^{-bnx}\ (e^{iax}-e^{-iax})dx=\frac{1}{2}\sum_{n=1}^{\infty }[\int_{0}^{\infty }x\ e^{-(bn-ia)}dx+\int_{0}^{\infty }x\ e^{-(bn+ia)}dx]\\ \\ \\ =\frac{1}{2}\sum_{n=1}^{\infty }(\frac{\Gamma (2)}{(bn-ai)^2}+\frac{\Gamma (2)}{(bn+ia)^2})=\frac{1}{2b^2}\sum_{n=0}^{\infty }\frac{1}{(n-\frac{ai}{b})^2}+\frac{1}{2b^2}\sum_{n=0}^{\infty }\frac{1}{(n+\frac{ai}{b})^2}+\frac{1}{a^2}\\ \\$$ $$=\frac{1}{a^2}+\frac{1}{2b^2}(\Psi ^{1}(\frac{ai}{b})+\Psi ^{1}(\frac{-ai}{b}))\\\\\ \\ but\ we\ know\ \Psi ^{(1)}(\frac{-ai}{b})=\Psi ^{(1)}(1-\frac{ai}{b})-\frac{b^2}{a^2}\\ \\ \\ \therefore \ I=\frac{1}{a^2}+\frac{1}{2b^2}\left ( \Psi ^{(1)}(1-\frac{ai}{b}) +\Psi ^{(1)}(\frac{ai}{b})-\frac{b^2}{a^2}\right )\\ \\ \\ by\ using\ the\ reflection\ formula\ :\ \Psi ^{(1)}(1-\frac{ai}{b})+\Psi ^{(1)}(\frac{ai}{b})=\frac{\pi ^2}{sin^2(\frac{i\pi a}{b})}\\ \\$$

so we have $$\therefore I=\frac{1}{2a^2}+\frac{1}{2b^2}\left ( \frac{-\pi ^2}{sinh^2(\frac{\pi a}{b})} \right )\\ \\ \\ =\frac{1}{2a^2}-\frac{\pi ^2}{2b^2sinh^2(\frac{\pi a}{b})}\ \ \ \ \ \ , b>0$$

note that : $$\frac{\pi ^2}{sin^2(\frac{i\pi a}{b})}=-\frac{\pi ^2}{sinh^2(\frac{\pi a}{b})}$$