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Evaluate $$\int_{0}^{\infty}\frac{\log^2 x}{e^{x^2}}\mathrm{d}x$$.

EDIT

Thank you for putting the question on hold and leaving me without any idea. Now that I have solved it I think it's a quite nice integral and so I'm glad I managed to do it on my own. Here is my solution.

Exponent with negative argument in integral from $0$ to $\infty$ reminds us of the gamma function. Recall the definition: $\Gamma(s)=\int_{0}^{\infty}e^{-x}x^{s-1}\mathrm{d}x$. In this way the logarithm comes quite naturally as we take derivative of $\Gamma$ using diferentiation under the integral sign. Second power of the $\log$ comes from second derivative of $\Gamma$. The situation is a bit more complicated here, since we have $e$ to the power $x^2$. However this is not a great obstacle - just make the change $x\to\sqrt{x}$. Thus $\int_{0}^{\infty}e^{-x^2}x^{s-1}\mathrm{d}x=\int_{0}^{\infty}e^{-x}x^{\frac{s-1}{2}}\frac{1}{2\sqrt{x}}\mathrm{d}x=\frac{1}{2}\Gamma(\frac{s}{2})$. Hence $$\int_{0}^{\infty}e^{-x^2}\log^2 x =\frac{\mathrm{d}^2}{\mathrm{d}s^2}\Big|_{s=1}\int_{0}^{\infty}e^{-x^2}x^{s-1}\mathrm{d}x=\frac{1}{8}\Gamma''\left(\frac{1}{2}\right).$$ The problem now reduces to finding second derivative of gamma at $\frac{1}{2}$. To find derivatives of $\Gamma$ it is helpful to find derivatives of $\log\Gamma$ - the so-called polygamma function. Representing $\Gamma$ as its Weierstrass factorization and taking $\log$ leads to series which are simple to differentiate and calculate. Recall that $$\Gamma(s)=\frac{e^{-\gamma s}}{s}\prod_{n\ge 1}^{\infty}\left(\left(1+\frac{s}{n}\right)^{-1} e^{s/n}\right).$$ Thus $$\frac{\mathrm{d}}{\mathrm{d}s}\Big|_{s=\frac{1}{2}}\log\Gamma(s)=\frac{\mathrm{d}}{\mathrm{d}s}\Big|_{s=\frac{1}{2}}\left(-\gamma s-\log s+\sum_{n\ge 1}^{\infty}\left(-\log\left(1+\frac{s}{n}\right)+\frac{s}{n}\right)\right)=-\gamma-\frac{1}{s}+\sum_{n\ge 1}^{\infty}\left(-\frac{1}{n+s}+\frac{1}{n}\right)\Big|_{s=\frac{1}{2}}=-\gamma-2+2\sum_{n=1}^{\infty}\frac{1}{2n(2n+1)}=-\gamma+\log 4.$$ Now rewrite $\frac{\mathrm{d}}{\mathrm{d}s}\Big|_{s=\frac{1}{2}}\log\Gamma(s)=\frac{\Gamma'\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right) }$. So we have $$\Gamma'\left(\frac{1}{2}\right)=\Gamma\left(\frac{1}{2}\right)(-\gamma+\log 4)\ (1).$$ In the same manner we find second derivative of $\log\Gamma$ $$\frac{\mathrm{d}^2}{\mathrm{d}s^2}\Big|_{s=\frac{1}{2}}\log\Gamma(s)=\frac{1}{s^2}+\sum_{n\ge 1}^{\infty}\frac{1}{(n+s)^2}\Big|_{s=\frac{1}{2}}=4+4\sum_{n=1}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{2}.$$ Again we rewrite the derivative as differentiation of composite function - $\frac{\mathrm{d}^2}{\mathrm{d}s^2}\Big|_{s=\frac{1}{2}}\log\Gamma(s)=-\frac{\Gamma'\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)^2}+\frac{\Gamma''\left(\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)}$. Now, taking into consideration what we previously obtained about the second derivative, the result about the first derivative and the fact $\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$ we come to equation for the second derivative at $\frac{1}{2}$ which gives us $$\Gamma''\left(\frac{1}{2}\right)=\frac{\sqrt{\pi}}{2}\left(2\gamma^2+\pi^2+4\gamma \log 4+2\log^2 4\right).$$ Now mupltiply by $8$ and get the final result.

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    $\begingroup$ i know the answer but won't tell it to you until you have shown some own efforts...:p $\endgroup$ – tired Jan 7 '17 at 13:42
  • $\begingroup$ See here. $\endgroup$ – Rohan Jan 7 '17 at 13:45
  • $\begingroup$ Now, now, @tired, one can ask for effort shown without having to be a tease... :) $\endgroup$ – J. M. is a poor mathematician Jan 7 '17 at 13:46
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    $\begingroup$ @J.M.isn'tamathematician Teasing is fun from time to time ^^ $\endgroup$ – tired Jan 7 '17 at 13:47
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    $\begingroup$ (+1) Don't get sad, :) this happens a lot that some good questions get closed. As an advice always try to put some motivation for the question or mention some of your thoughts for solving it. Good Luck. :) $\endgroup$ – H. R. Jan 7 '17 at 19:28
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Consider

$$I(a) = \int_0^{\infty} dx \, x^a\, e^{-x^2} = \frac12 \int_0^{\infty} dy \, y^{(a-1)/2} \, e^{-y} = \frac12 \Gamma \left ( \frac{a+1}{2} \right )$$

The sought-after integral is $I''(0)$, where

$$\begin{align}I''(a) &= \frac14 \frac{d}{da} \left [ \Gamma \left ( \frac{a+1}{2} \right ) \psi \left ( \frac{a+1}{2} \right ) \right ] \\ &= \frac18 \left [ \Gamma \left ( \frac{a+1}{2} \right )^2 \psi \left ( \frac{a+1}{2} \right ) + \Gamma \left ( \frac{a+1}{2} \right ) \psi_1 \left ( \frac{a+1}{2} \right )\right ]\\ I''(0) &= -\frac{\pi}{8} (2 \log{2} + \gamma) + \frac{\pi^{5/2}}{16}\end{align}$$

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