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″(12).
The problem now reduces to finding second derivative of gamma at 12. To find derivatives of Γ it is helpful to find derivatives of logΓ - the so-called polygamma function. Representing Γ as its Weierstrass factorization and taking log leads to series which are simple to differentiate and calculate. Recall that Γ(s)=e−γss∞∏n≥1((1+sn)−1es/n). Thus dds|s=12logΓ(s)=dds|s=12(−γs−logs+∞∑n≥1(−log(1+sn)+sn))=−γ−1s+∞∑n≥1(−1n+s+1n)|s=12=−γ−2+2∞∑n=112n(2n+1)=−γ+log4. Now rewrite dds|s=12logΓ(s)=Γ′(12)Γ(12). So we have Γ′(12)=Γ(12)(−γ+log4) (1).
In the same manner we find second derivative of logΓ
d2ds2|s=12logΓ(s)=1s2+∞∑n≥11(n+s)2|s=12=4+4∞∑n=11(2n+1)2=π22.
Again we rewrite the derivative as differentiation of composite function - d2ds2|s=12logΓ(s)=−Γ′(12)Γ(12)2+Γ″(12)Γ(12). Now, taking into consideration what we previously obtained about the second derivative, the result about the first derivative and the fact Γ(12)=√π we come to equation for the second derivative at 12 which gives us Γ″(12)=√π2(2γ2+π2+4γlog4+2log24). Now mupltiply by 8 and get the final result.
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