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2 integrals, one easy, one hard

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2 integrals, one easy, one hard

Wed Aug 10, 2016 5:36 pm

Posts: 38
Easy one first. Show that

$$\int_{-\infty}^{\infty}\log(1+e^{-x^{2}})\text{d}x=\left(1-\frac{\sqrt{2}}{2}\right)\zeta\left(\frac{3}{2}\right)\sqrt{\pi}$$

The other one, I'm not sure if there is a closed form (and I wouldn't be surprised if there isn't one). Evaluate

$$I=\int_{e}^{e^{e}}\log\log\log x\,\text{d}x$$

I've given it a short attempt and the best I got was

$$I=(e-1)\gamma-\text{PV}\int_{-\infty}^{1}\frac{e^{e^{x}}-1}{x}\text{d}x+\sum_{n=2}^{\infty}\frac{\log n}{n!}$$

where PV is the principal value, given by

$$\text{PV}\int_{-\infty}^{1}\frac{e^{e^{x}}-1}{x}\text{d}x=\lim_{\epsilon\to 0}\left(\int_{-\infty}^{-\epsilon}+\int_{\epsilon}^{1}\right)\frac{e^{e^{x}}-1}{x}\text{d}x$$

Re: 2 integrals, one easy, one hard

Fri Aug 12, 2016 12:19 am

Posts: 46
Location: http://www.tolaso.com.gr/
ben1996123 wrote:
Easy one first. Show that

$$\int_{-\infty}^{\infty}\log(1+e^{-x^{2}})\text{d}x=\left(1-\frac{\sqrt{2}}{2}\right)\zeta\left(\frac{3}{2}\right)\sqrt{\pi}$$

\begin{align*}
\int_{-\infty}^{\infty} \log \left ( 1+e^{-x^2} \right ) \, {\rm d}x &=2\int_{0}^{\infty} \log \left ( 1+e^{-x^2} \right ) \, {\rm d}x \quad \text{(due to parity)} \\
&= 2\int_{0}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1} e^{-nx^2}}{n} \, {\rm d}x\\
&= 2\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \int_{0}^{\infty} e^{-n x^2} \, {\rm d}x\\
&\overset{(\dagger)}{=} \sqrt{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n \sqrt{n}} \\
&\!\overset{(\dagger \dagger)}{=\! =\!}\sqrt{\pi} \left ( 1- \frac{\sqrt{2}}{2} \right ) \zeta \left ( \frac{3}{2} \right ) \\
\end{align*}

$(\dagger)$ Easy to prove that $\displaystyle \int_{0}^{\infty} e^{-nx^2} \, {\rm d}x = \frac{\sqrt{\pi}}{2\sqrt{n}}$. (It is actually a shifted Gaussian)

$(\dagger \dagger)$ Also easy to prove that $\displaystyle \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n \sqrt{n}}= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{3/2}}=\left ( 1-\frac{\sqrt{2}}{2} \right ) \zeta \left ( \frac{3}{2} \right )$. (We are using the $\eta$ Dirichlet function and the functional equation it satisfies

$$\eta(s)=\left(1 -2^{1-s} \right) \zeta(s)$$

and the result follows.)