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I was playing around with the integral: $$\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz$$

I couldn't find a way of solving it, but I used WolframAlpha to find that the integral evaluated to $\zeta(2)$ or $\frac{\pi^2}{6}$.

How would I go around finding the solution?

My experience with integrals is very limited, but I think the solution can possibly be found using the polylogarithm (saw a similar integral use it), but I don't know much of the function so maybe someone on here could help me.

I initially tried rewriting the $\frac{1}{1+e^{-z}}$ as a geometric series, but I don't think that works.

Then I looked at finding the poles and drawing a contour, but I also don't think that works.

Is there some identity that can be used here, which I happen not to know?

Any help is appreciated, especially if you could show your working out step by step.

Gary
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Abdullah
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3 Answers3

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My immediate thought is to let $$ u = 1+e^{-z} \implies du = -e^{-z} dz = (1-u)dz $$ so $$ \mathcal{I}:= \int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz = \int_{\infty}^{1} \frac{\log(u)}{u} \frac{du}{1-u} = \int_{1}^{\infty} \frac{\log(u)}{u(u-1)} du $$ Now let $$ v = \log(u) \implies dv = \frac 1 u du $$ giving us a perhaps more familiar integral: $$ \mathcal{I} = \int_{0}^{\infty} \frac{v}{ e^v-1 } dv = \int_{0}^{\infty} \frac{ve^{-v}}{ 1-e^{-v} } dv $$ Apply the geometric series to this: assuming the interchange of sum & integral is valid, $$ \mathcal{I} = \int_{0}^{\infty} ve^{-v} \sum_{n=0}^\infty e^{-nv} dv = \sum_{n=0}^\infty \int_{0}^{\infty} ve^{-v} e^{-nv} dv = \sum_{n=0}^\infty \int_{0}^{\infty} v e^{-(n+1)v} dv = \sum_{n=1}^\infty \int_{0}^{\infty} v e^{-nv} dv $$ The remainder is a standard integral. Apply integration by parts. $$ \mathcal{I} = \sum_{n=1}^\infty \frac{1}{n^2} = \zeta(2) $$

PrincessEev
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  • Thank you! Didn't think it would be as simple as this. Would this also work if all the $e^{-z}$s were $e^{z}$? And is there a way of generalizing this sort of integral for $ζ(2n)$, just some extra thoughts I've had relating to this. But again, thanks for answering. – Abdullah Jan 22 '24 at 09:27
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Substitute $t=\frac1{1+e^{-z}}$ and then utilize $\int_0^1 \frac{\ln x}{x-1}dx=\frac{\pi^2}6 $ to obtain

$$\int_{-\infty}^{\infty} \frac{\log(1+e^{-z})}{1+e^{-z}}dz=\int_0^1 \frac{\ln t}{t-1}dt=\frac{\pi^2}6$$

Quanto
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$$ I \stackrel{y=1+e^{-z}}{=} \int_1^{\infty} \frac{\log y}{y(1-y)} d y \stackrel{y\mapsto\frac{1}{y}}{=} -\int_0^1 \frac{\log y}{1-y} d y= -\sum_{n=0}^{\infty} \int_0^1 y^n \log y d y \stackrel{IBP}{=} \zeta(2)=\frac{\pi^2}{6} $$

Lai
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