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I'm looking for a method to evaluate the following integral:

$\displaystyle \int_0^{\infty} \left( \frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) \frac{1}{x} dx$

EDIT:

Using the link, here's what I've done so far. Define

$F(s) = \displaystyle \int_0^{\infty} e^{-sx} \left (\frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) \frac{1}{x} dx$.

Supposing we've shown that this integral converges nicely enough to justify differentiating underneath it, we have that

$F'(s) = \displaystyle -\int_0^{\infty} e^{-sx} \left(\frac{1}{e^x - 1} - \frac{1}{x} + \frac{e^{-x}}{2} \right) dx = - \left( \int_0^{\infty}e^{-sx} \frac{e^{-x}}{2} dx + \int_0^{\infty} e^{-sx} \left( -\frac{1}{x} + \frac{1}{e^x - 1} \right) dx \right)$.

We recognize that the first integral is the Laplace transform of $\frac{e^{-x}}{2}$ and so the integral evaluates to $\frac{1}{2(s+1)}$.

Setting $F_2(s) \displaystyle \int_0^{\infty} e^{-sx} \left(-\frac{1}{x} + \frac{1}{e^{x} - 1} \right) dx$, (and assuming a similar analysis will justify the differentiation), we find that

$F'_2(s) = \displaystyle \int_0^{\infty} e^{-sx} - \frac{x e^{-sx}}{e^x - 1} dx = \frac{1}{s} - \int_0^{\infty} \frac{xe^{-sx}}{e^x - 1} dx $.

While I'm pretty sure I can justify the convergence of all the relevant integrals to differentiate, this last integral is not the Laplace transform of a function I'm familiar with (which doesn't mean much, because I'm not familiar with many). Any tips on how to take care of the last integral?

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  • $\begingroup$ Do you mean that the expression in divided by x or $d(\frac xx)$ $\endgroup$ – superAnnoyingUser Mar 3 '13 at 19:40
  • $\begingroup$ The entire expression is divided by x. I've changed the notation to reduce ambiguity. $\endgroup$ – anonymous Mar 3 '13 at 19:42
  • $\begingroup$ Ah. I did not see that at all, deleted my first response. $\endgroup$ – muzzlator Mar 3 '13 at 19:43
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    $\begingroup$ A related problem. $\endgroup$ – Mhenni Benghorbal Mar 3 '13 at 19:48
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$$f(\lambda)=\int_0^{\infty} \frac{e^{-\lambda x}}{x}\left(\frac{1}{e^x-1}+\frac{1}{2e^x}-\frac{1}{x}\right)dx$$

$$\begin{align*}f''(\lambda)&=\int_0^{\infty} e^{-\lambda x}\left(\frac{x}{e^x-1}+\frac{x}{2e^x}-1 \right)dx\\&=\int_0^{\infty} \frac{x}{e^{\lambda x}(e^x-1)}dx+\int_0^{\infty}\frac{x}{2e^{(\lambda +1)x}}dx-\int_0^{\infty} e^{-\lambda x} dx\\&=\int_0^{\infty} \frac{x}{e^{\lambda x}(e^x-1)}dx+\frac{1}{2(\lambda+1)^2}-\frac{1}{\lambda}\end{align*}$$

Recall the Hurwitz zeta function:

$$\zeta (s,w)=\frac{1}{\Gamma (s)}\int_0^{\infty}\frac{t^{s-1}}{e^{wt}(1-e^{-t})}dt$$

So

$$f''(\lambda)=\zeta(2,\lambda+1)+\frac{1}{2(\lambda+1)^2}-\frac{1}{\lambda}$$

Now integrate back twice:

$$\int f''(\lambda)d\lambda =\frac{d}{d\lambda}\ln\Gamma (\lambda+1)-\frac{1}{2(\lambda+1)}-\ln \lambda+\mathcal{C}$$

Letting $\lambda\to\infty$ reveals that $\mathcal{C}=0$

$$\int f'(\lambda)d\lambda =\ln\Gamma (\lambda+1)+\lambda (1-\ln\lambda)-\frac{1}{2}\ln (\lambda +1)+\mathcal{C'}$$

Taking $\lambda\to\infty$ we have $\mathcal{C'}=\ln\sqrt{\dfrac{1}{2\pi}}$ (easily deduced using Stirling's approximation)

Now letting $\lambda\to 0$ gives us the result:

$$\int_0^{\infty} \frac{1}{x}\left(\frac{1}{e^x-1}+\frac{1}{2e^x}-\frac{1}{x}\right)dx=\ln\sqrt{\dfrac{1}{2\pi}}$$

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