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Evaluate the integral $$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx$$

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I started by stating that the integral from 0 to infinity should be the same as half the integral from $-\infty$ to $\infty$, that is:

$$\int_0^\infty \frac{\log(2+x^2)}{4+x^2}dx = \frac{1}{2}\int_{-\infty}^\infty \frac{\log(2+x^2)}{4+x^2}dx$$

and then by stating that this must be equal to

$$\pi i \cdot \sum(\text{residues in upper plane})$$

noting that there is a singularity at $z=2i$ in the upper half plane, and an "issue" (I don't know if it's strictly a singularity) with the log function at $z=\sqrt{2}i$.

The residue at $z=2i$ is easily dealt with, but when I try to deal with the log issue, I can't make any headway. I decided to make my branch cut between $z=\sqrt{2}i$ and $z=-\sqrt{2}i$, and form a contour that goes around this cut and the point, but it doesn't seem to be working out for me.

Suggestions would be appreciated!

Maximilian Janisch
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MonsterEater
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  • https://math.stackexchange.com/questions/2138915/improper-integral-of-frac-log-left-sqrtx2a2-rightx2b2/2140045#2140045 – Ron Gordon Mar 12 '20 at 09:12

3 Answers3

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Using Feynman's trick works:

For $a\geq0$, let $$G(a)=\int_0^\infty \frac{\ln(2+ ax^2)}{4+x^2}\,\mathrm dx.$$

Then the Leibniz rule implies that for $a>0$, $$G'(a)=\int_0^\infty \frac{x^2}{\left(x^2+4\right) \left(a x^2+2\right)}\,\mathrm dx.$$ The last integrand is a rational function and thus has a closed form anti-derivative that can be computed algortihmically. Namely,

$$G'(a)=\left. \frac{\frac{\sqrt{2} \tan ^{-1}\left(\frac{\sqrt{a} x}{\sqrt{2}}\right)}{\sqrt{a}}+2 \tan ^{-1}\left(\frac{2}{x}\right)}{2-4 a}\right|_{x=0}^{x=\infty}=\dots=\frac{\pi}{2\sqrt2(1+\sqrt2\sqrt a)\sqrt a}.$$

We know that $G(a)=G(0)+\int_0^a G'(b)\,\mathrm db$. Indeed, $$\int_0^a G'(b)\,\mathrm db = \frac{\pi}{2\sqrt2} \int_0^a \frac1{(1+\sqrt{2}\sqrt b)\sqrt b}\,\mathrm db=\frac\pi2\ln(1+\sqrt2\sqrt a).$$

Also, $$G(0)=\ln(2)\int_0^\infty \frac{1}{4+x^2}\,\mathrm dx=\frac{\ln(2)\pi}4.$$

Hence,

$$\bbox[15px,border:1px groove navy]{ \int_0^\infty \frac{\ln(2+ x^2)}{4+x^2}\,\mathrm dx =G(1)=\frac{\pi}2\left(\frac{\ln(2)}2+\ln(1+\sqrt2)\right). }$$

Ali Shadhar
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Maximilian Janisch
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2

$$I=\int_{0}^{\infty} \frac{\log(2+x^2)}{4+x^2}dx=\int_{0}^{\infty} \frac{\log(a^2+x^2)}{4+x^2}dx=I(a)$$ $$\implies \frac{dI}{da}=2a \int_{0}^{\infty} \frac{dx}{(a^2+x^2)(4+x^2)]}= \frac{2a}{4-a^2}\int_{0}^{\infty} \left( \frac{1}{a^2+x^2}-\frac{1}{4+x^2}\right) dx$$ $$\frac{dI}{da}=\frac{2a}{4-a^2}[\frac{\pi}{2a}-\frac{\pi}{4}]=\frac{\pi}{2(a+2)} \implies I(a)=\frac{\pi}{2} \ln (a+2)+C~~~~(*)$$ $$I(0)=2\int_{0}^{\infty} \frac{\ln x}{x^2+4}dx=J~~~~(1)$$ Let $x=1/t \implies dx=-dt/t^2$ in $J$, the $$J=2\int_{0}^{\infty} \frac{-\ln t}{t^2+4}dt=2\int_{0}^{\infty} \frac{-\ln x}{x^2+4}dx~~~~(2)$$ Adding (1) and (2), we get $J=0=I(0)$, from (*) we get $C=-\frac{\pi}{2}\ln 2$ Thus, $$I(a)=\frac{\pi}{2} \ln (a+2)-\frac{\pi}{2} \ln 2= \frac{\pi}{2}\ln\frac{a+2}{2}=\frac{\pi}{2}\ln (1+\frac{1}{\sqrt{2}}),~ as ~a=\sqrt{2}.$$

Z Ahmed
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Without residues. $$ \frac{\log(2+x^2)}{4+x^2}=\frac{\log(x+i\sqrt2)+\log(x-i\sqrt2)}{(x+2i)(x-2i)}$$ $$\frac{1}{(x+2i)(x-2i)}=\frac{i}{4}\left(\frac{1}{x+2 i}-\frac{1}{x-2 i}\right)$$ So, we face four integrals looking like $$\int \frac{\log(x+a)}{x+b}\,dx=\text{Li}_2\left(\frac{x+a}{a-b}\right)+\log (x+a) \log \left(\frac{b+x}{b-a}\right)$$

Claude Leibovici
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