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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
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
Evaluate $$\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx$$
Hint: Let $$A(a)=\displaystyle{\int_0^{\infty}\frac{\ln (1+ax^2)}{1+x^2}dx}$$
04-20-2010, 10:07 AM
Post: #2
 elim Moderator Posts: 578 Joined: Feb 2010 Reputation: 0
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 $\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx=A(1)=\pi \ln 2$
09-20-2012, 06:16 AM
Post: #3
 L.Hartsdale L.Hartsdale Posts: 1 Joined: Sep 2012 Reputation: 0
RE: Evaluate $$\int_0^{\infty}\left(\ln (x^2+1)/(x^2+1) \right)dx$$
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