Is there any closed-form representation for the following integral? $$\int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx,$$ where $\mathrm{sech}\,x$ is the hyperbolic secant, $\mathrm{sech}\,x=\dfrac{2}{e^x+e^{-x}}$.
Asked
Active
Viewed 585 times
18
-
I wonder if mathematica 9 gives a closed form for this integral. – Mhenni Benghorbal May 21 '13 at 16:54
-
1@MhenniBenghorbal No, Mathematica 9 cannot solve it. – Oksana Gimmel May 23 '13 at 03:02
-
Nor can Mathematica 12. But interestingly, it can solve $\int_0^\infty(\log x)(\operatorname{sech} x)^2dx.$ – WillG Aug 04 '21 at 23:27
2 Answers
12
Hint: Use the integral: $$\int_0^\infty x^n\,\text{sech}^2x\,\mathrm dx=(2^{1-n}-4^{1-n})\,\Gamma(n+1)\,\zeta(n)$$ and take the 2nd derivative with respect to $n$.
Vladimir Reshetnikov
- 47,122
-
-
Actually, I figured it out thanks to the long hint given here: https://math.stackexchange.com/questions/464073/show-that-int-0-infty-xn-textsech2x-mathrm-dx-21-n-41-n-gam – WillG Aug 05 '21 at 07:34
3
Related technique: (I), (II), (III). Here is a closed form
$$ \int_0^\infty(\log x)^2(\mathrm{sech}\,x)^2\mathrm dx = \int_0^\infty(\log x)^2\left(\frac{2}{e^{x}+e^{-x}}\right)^2\mathrm dx $$
$$= \left( \ln \left( \pi \right) \right) ^{2}- \left( 4\,\ln \left( 2 \right) +2\,\gamma \right) \ln \left( \pi \right) +2\, \left( \ln \left( 2 \right) \right) ^{2}+4\,\ln \left( 2 \right) \gamma-2\,\gamma \left( 1 \right) +\frac{1}{4}\,{\pi }^{2} $$
$$ \sim 1.989349759. $$
Note: $\gamma(n)$ are known as Stieltjes $\gamma$-constants
$$ \gamma(n)= \lim_{m\to \infty}\left(\sum_{k=1}^m \frac{(\ln k)^n}{k}-\frac{(\ln m)^{n+1}}{n+1}\right).$$
Mhenni Benghorbal
- 47,431