Evaluate ∫∞0log2xex2dx.
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 ∞ reminds us of the gamma function. Recall the definition: Γ(s)=∫∞0e−xxs−1dx. In this way the logarithm comes quite naturally as we take derivative of Γ using diferentiation under the integral sign. Second power of the log comes from second derivative of Γ. The situation is a bit more complicated here, since we have e to the power x2. However this is not a great obstacle - just make the change x→√x. Thus ∫∞0e−x2xs−1dx=∫∞0e−xxs−1212√xdx=12Γ(s2). Hence ∫∞0e−x2log2x=d2ds2|s=1∫∞0e−x2xs−1dx=18Γ″
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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