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Evaluate \(\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx\)
04-09-2010, 03:04 PM
Post: #1
Evaluate \(\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx\)
Hint: Let \(A(a)=\displaystyle{\int_0^{\infty}\frac{\ln (1+ax^2)}{1+x^2}dx}\)
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04-20-2010, 10:07 AM
Post: #2
RE: Evaluate \(\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx\)
Define \(A(a)=\displaystyle{\int_0^{\infty} \frac{\ln(1+ax^2)}{1+x^2}dx} \), then
\(A'(a)=\displaystyle{\int_0^{\infty} \frac{x^2}{(1+ax^2)(1+x^2)}dx}= (1-a)^{-1}\int_0^{\infty}\left( (ax^2+1)^{-1} - (1+x^2)^{-1}\right )dx\)
\(=\displaystyle{\frac{\pi}{2}\left(\frac{1}{\sqrt{a}(1-a)}-\frac{1}{1-a} \right)}=\frac{d}{da}(\pi \ln(\sqrt{a}+1))\)
Since \(A(0)=0\) we see that \(A(a)=\pi \ln(\sqrt{a}+1)\) and so \[\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx=A(1)=\pi \ln 2 \]
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09-20-2012, 06:16 AM
Post: #3
RE: Evaluate \(\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx\)
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