Hey I am trying to integrate $$ \int_0^\infty \frac{x^n\ln x}{(x^2+\alpha^2)^2(e^x-1)}dx,\quad \alpha,n \in \mathbb{R}^{0+}. $$ This integral is old. I am also looking for literature on these integrals as I have seen many for small values of n, and variations of this. Thanks. Maybe we can use residues however the log function in the denominator is what is concerning me, without that I can see what to do

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    $\begingroup$ @Lucian ok, and...? How do you suppose to do that integral? I am looking for a proof, not comments about standard substitutions involving beta functions. Thanks. $\endgroup$ Commented Apr 11, 2014 at 22:43
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    $\begingroup$ Do we know the answer for $n=1$? Perhaps we can use Ron's post again - residuetheorem.com/2014/01/15/… but considering the integral $$ \oint_C dz \frac{z \log^2{z}}{(e^z-1)(z^2+\alpha^2)^2} $$ around the same contour? $\endgroup$ Commented May 6, 2014 at 2:24
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    $\begingroup$ If there exists a formula for $$ \int_0^\infty \frac{x^n}{(x^2+\alpha^2)^2(e^x-1)}dx,\quad \alpha,n \in \mathbb{R}^{0+}, $$ differentiating w.r.t. $n$ would give us the result, no? Alas I'm not sure this is doable with a general $\alpha$, given that $\alpha=2n\pi$ is a special case of the triple pole. $\endgroup$ Commented May 6, 2014 at 3:40
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    $\begingroup$ @BennettGardiner that is correct, I agree with you. Interesting observation! Yes that is the special case. Thank you for taking the time to care about integrals that arise in nature $\endgroup$ Commented May 6, 2014 at 4:17
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    $\begingroup$ @BennettGardiner Well I do not mean it arises in engineering or physics. I meant an integral in the mathematical nature, the mathematical beauty, sorry :) I am not sure if it is in any physic books $\endgroup$ Commented May 6, 2014 at 4:22

1 Answer 1


This is inspired by the game that @BennettGardiner played in this question with Binet's second Log-Gamma integral formula: $$ \ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t $$

In the above, shift the integration variable $t$ to the coordinate $x=2\pi t$, and then define the constant $\alpha:= 2 \pi z$. Rearranging the terms, we are left with the following integral: $$ \int_{0}^{\infty} \frac{\tan^{-1}\left( \tfrac{x}{\alpha} \right)}{e^{x}-1} dx \ = \ \pi \ln\left[ \Gamma\left(\frac{\alpha}{2\pi}\right) \right] - \frac{\alpha - \pi}{2} \ln\left( \frac{\alpha}{2\pi} \right) + \frac{\alpha}{2} - \frac{\pi}{2} \ln(2\pi) $$

Now noting that $\frac{d}{d\alpha}\left\{ \tan^{-1}\left( \tfrac{x}{\alpha} \right) \right\} = - \frac{x}{x^2 + \alpha^2}$, we hit both sides of the above relation with $-\frac{d}{d\alpha}$ and get the following: $$ \int_{0}^{\infty} \frac{x}{(e^{x}-1)(x^2+\alpha^2)} dx \ = \ \frac{1}{2} \ln\left( \frac{\alpha}{2 \pi} \right) - \frac{\pi}{2 \alpha} - \frac{1}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right) $$ where $\psi^{(0)}$ is the digamma (aka polygamma-0) function. This is the generalization of the integral from that same question I mentioned above.

Next we take note of the following derivative: $$ \frac{d}{d\alpha} \left\{ \frac{x}{x^2 + \alpha^2} \right\} = - \frac{2 \alpha x}{(x^2 + a^2)^2} $$

So hitting both sides of the above integral with the operator $- \frac{1}{2\alpha} \frac{d}{d\alpha}$ yields the following integral: $$ \int_{0}^{\infty} \frac{x}{(e^{x}-1)(x^2+\alpha^2)^2} dx \ = \ - \frac{1}{4\alpha^2} - \frac{\pi}{4 \alpha^3} + \frac{1}{8 \pi \alpha} \psi^{(1)}\left( \frac{\alpha}{2\pi} \right) $$

where $\psi^{(1)}$ is the polygamma-$1$ function.

Next, we note the following: $$ \frac{d}{d\alpha} \left\{ \frac{\alpha^2 x}{x^2+\alpha^2} \right\} = \frac{2 \alpha x^3}{(\alpha^2 + x^2)^2} $$

So we take our second integral identity, multiply both sides by $\alpha^2$ and then hit both sides with $\frac{d}{d\alpha}$: $$ \frac{d}{d\alpha} \left\{ \int_{0}^{\infty} \frac{x\alpha^2}{(e^{x}-1)(x^2+\alpha^2)} dx \right\} \ = \ \frac{d}{d\alpha} \left\{ \frac{\alpha^2}{2} \ln\left( \frac{\alpha}{2 \pi} \right) - \frac{\pi \alpha}{2} - \frac{\alpha^2}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right) \right\} $$

Taking the derivative in the above, and then dividing everything by $2\alpha$ we are left with the formula: $$ \int_{0}^{\infty} \frac{x^3}{(e^{x}-1)(x^2+\alpha^2)^2} dx \ = \ \frac{1}{4} - \frac{\pi}{4 \alpha} + \frac{1}{2} \log\left( \frac{\alpha}{2\pi} \right) - \frac{1}{2} \psi^{(0)}\left( \frac{\alpha}{2\pi} \right) - \frac{\alpha}{8 \pi} \psi^{(1)}\left( \frac{\alpha}{2\pi} \right) $$

In summary, I have found two integrals related to your integral (without the log term though) for $n=1,3$. I would like to find a general formula for any $n \geq 0$ so we could apply an operator $\frac{d}{dn}$ to get down a $\ln(x)$ as @BennettGardiner suggested above. I don't know where to go from here though.


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