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Evaluate $$ I=\int_0^\infty \frac{\log (1+x^2)}{1+x^2} dx $$

I tried several 'obvious' substitutions e.g. $y=1+x^2, y=\log(1+x^2)$, tried integrating by parts, and combinations of substitutions and integrating by parts, but none of these approaches seemed to work out.

Finally, I have achieved this as follows. By first substituting $x=\tan \theta$ to give

$$I= -2\int_0^{\pi/2} \log \cos \theta\, d\theta$$

Then using the property that $$I= \int_0^{\pi/2} f( \cos \theta) \, d\theta = \int_0^{\pi/2} f( \sin \theta) \, d\theta$$ to obtain another expression for $I$ $$I= -2\int_0^{\pi/2} \log \sin \theta\, d\theta$$

Aadding these two expressions for $I$ yields (using $\log A + \log B = \log AB$) $$ -I = \int_0^{\pi/2} \log \sin 2 \theta\, d\theta - \frac{\pi}{2} \log2$$

Now substitute $t=2 \theta$ $$ -I = \frac{1}{2}\int_0^{\pi} \log \sin t\, dt - \frac{\pi}{2} \log2 \tag{1}\label{1}$$

Then, when $f(2a-x) = f(x)$, use the property that $$ \int_0^{2a} f(x)\,dx = 2 \int_0^{a} f(x)\,dx$$ to get $$ \int_0^\pi \log \sin t \, dt = 2 \int_0^{\pi/2} \log \sin t \, dt = -I$$

Using this with equation $\ref{1}$ finally yields $$I = \pi \log2$$

This seems a quite long winded solution and I feel I have missed something much more straightforward. I would be very interested in simpler ways of doing this integral, if possible.

PM.
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1 Answers1

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In addition to the above method, we can also use integrals with parameters. Let \begin{align*} I(t)=\int_0^\infty\frac{\ln\left(1+tx^2\right)}{1+x^2}dx \end{align*} so we only need to solve for the value of $I(1)$. Derivating the above equation we get \begin{align*} I'(t)=\int_0^\infty \frac{x^2}{\left(1+tx^2\right)\left(1+x^2\right)}dx=\frac{\pi}{2\left(1+\sqrt{t}\right)\sqrt{t}} \end{align*} notice that $I(0)=0$ which determines the constant in indefinite integral, then \begin{align*} I(t)=\int\frac{\pi}{2\left(1+\sqrt{t}\right)\sqrt{t}}dt=\pi\ln\left(1+\sqrt{t}\right) \end{align*} finally we get $I(0)=\pi\ln{2}$.

The above procedure omits the process of solving some simple indefinite integrals, which I think you can solve easily.

Always
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