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\left(\dfrac{a\pi}{b}\right)}{2a b}=\frac12\left(-\frac{1}{a}+\frac\pi b\coth\left(\dfrac{a\pi}{b}\right)\right) \end{eqnarray} and hence $$ I'(a)=\frac12\left(\frac1{a^2}-\frac{\pi^2}{b^2}\text{csch}^2\left(\dfrac{a\pi}{b}\right)\right).$$

Answer 3

Let

$$ I = \int_{0}^{\infty} \frac{x \cos(ax)}{e^{bx} - 1} \, dx = \int_{0}^{\infty} \frac{x e^{-bx} \cos(ax)}{1 - e^{-bx}} \, dx, \quad \forall b > 0. $$

It is easy to see that:

$$ I = \sum_{n=1}^{\infty} \int_{0}^{\infty} x e^{-bnx} \cos(ax) \, dx = \frac{\partial}{\partial a} \sum_{n=1}^{\infty} \int_{0}^{\infty} e^{-bnx} \sin(ax) \, dx. $$

Thus, we have:

$$ I = \frac{\partial}{\partial a} \sum_{n=1}^{\infty} \frac{a}{a^{2} + b^{2} n^{2}}. $$

Recall that:

$$ \sum_{n=1}^{\infty} \frac{1}{x^{2} + n^{2}} = \frac{-1}{2x^{2}} + \frac{\pi (e^{x\pi} + e^{-x\pi})}{2x (e^{x\pi} - e^{-x\pi})}. $$

We can write:

$$ \sum_{n=1}^{\infty} \frac{a}{a^{2} + b^{2} n^{2}} = \frac{a}{b^{2}} \sum_{n=1}^{\infty} \frac{1}{\left(\frac{a}{b}\right)^{2} + n^{2}} = \frac{1}{a} \left( \left(\frac{a}{b}\right)^{2} \sum_{n=1}^{\infty} \frac{1}{\left(\frac{a}{b}\right)^{2} + n^{2}} \right). $$

Thus, we get:

$$ \sum_{n=1}^{\infty} \frac{a}{a^{2} + b^{2} n^{2}} = \frac{-1}{2a} + \frac{\pi}{2b} \coth\left(\frac{a\pi}{b}\right). $$

Taking the partial derivative of both sides, we conclude:

$$ \int_{0}^{\infty} \frac{x \cos(ax)}{e^{bx} - 1} \, dx = \frac{1}{2a^{2}} - \frac{\pi^{2}}{2b^{2}} \text{csch}^{2}\left(\frac{a\pi}{b}\right). $$

Answer 4

Let

$$ I = \int_{0}^{\infty} \frac{x \cos(ax)}{e^{bx} - 1} \, dx = \sum_{n=1}^{\infty} \int_{0}^{\infty} x \, e^{-bnx} \cos(ax) \, dx $$

$$ = \frac{1}{2} \sum_{n=1}^{\infty} \int_{0}^{\infty} x \, e^{-bnx} (e^{iax} - e^{-iax}) \, dx = \frac{1}{2} \sum_{n=1}^{\infty} \left[ \int_{0}^{\infty} x \, e^{-(bn - ia)x} \, dx + \int_{0}^{\infty} x \, e^{-(bn + ia)x} \, dx \right] $$

$$ = \frac{1}{2} \sum_{n=1}^{\infty} \left( \frac{\Gamma(2)}{(bn - ia)^2} + \frac{\Gamma(2)}{(bn + ia)^2} \right) = \frac{1}{2b^2} \sum_{n=0}^{\infty} \frac{1}{(n - \frac{ai}{b})^2} + \frac{1}{2b^2} \sum_{n=0}^{\infty} \frac{1}{(n + \frac{ai}{b})^2} + \frac{1}{a^2} $$

$$ = \frac{1}{a^2} + \frac{1}{2b^2} \left( \Psi^{(1)}\left(\frac{ai}{b}\right) + \Psi^{(1)}\left(-\frac{ai}{b}\right) \right) $$

But we know that

$$ \Psi^{(1)}\left(-\frac{ai}{b}\right) = \Psi^{(1)}\left(1 - \frac{ai}{b}\right) - \frac{b^2}{a^2} $$

Thus, we have

$$ I = \frac{1}{a^2} + \frac{1}{2b^2} \left( \Psi^{(1)}\left(1 - \frac{ai}{b}\right) + \Psi^{(1)}\left(\frac{ai}{b}\right) - \frac{b^2}{a^2} \right) $$

Using the reflection formula:

$$ \Psi^{(1)}\left(1 - \frac{ai}{b}\right) + \Psi^{(1)}\left(\frac{ai}{b}\right) = \frac{\pi^2}{\sin^2\left(\frac{i\pi a}{b}\right)} $$

We conclude that

$$ I = \frac{1}{2a^2} + \frac{1}{2b^2} \left( -\frac{\pi^2}{\sinh^2\left(\frac{\pi a}{b}\right)} \right) $$

$$ = \frac{1}{2a^2} - \frac{\pi^2}{2b^2 \sinh^2\left(\frac{\pi a}{b}\right)}, \quad b > 0 $$

Note that:

$$ \frac{\pi^2}{\sin^2\left(\frac{i\pi a}{b}\right)} = -\frac{\pi^2}{\sinh^2\left(\frac{\pi a}{b}\right)} $$