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(Motivation) This is an integral I made up for fun. WolframAlpha doesn't seem to come up with a closed form for it and I'm surprised there doesn't seem to be a duplicate after using Approach0, but I believe it equals

$$\pi\ln\left(\frac{5}{2}\right)$$

(Question) Since there seems to be a nice-looking closed form, could there be a cooler or more elegant way of solving this other than my attempt below?

If anyone is willing to look over my attempt and provide constructive criticism, I would greatly appreciate it as someone who seeks to expand his skills in complex analysis.

(Attempt) Let $f(z) = \displaystyle\frac{\log(1/2+z+z^2)}{1+z^2}$. Its poles are $z \in \left\{i,-i\right\}$ and its branch points are $\displaystyle z \in \left\{-\frac{1}{2}+\frac{i}{2}, -\frac{1}{2}-\frac{i}{2}\right\}$. For simplicity's sake, let $\displaystyle z_a = -\frac{1}{2}+\frac{i}{2}$ and $\displaystyle z_b = -\frac{1}{2}-\frac{i}{2}$. With these, we can rewrite $f(z)$ as $$ \begin{align} \frac{\log\left(\frac{1}{2}+z+z^{2}\right)}{1+z^{2}} &= \frac{\log\left(\left(z-z_{a}\right)\left(z-z_{b}\right)\right)}{1+z^{2}} \\ &= \frac{1}{1+z^{2}}\left(\log\left(\left|z-z_{a}\right|\cdot\left|z-z_{b}\right|\right)+i\operatorname{arg}\left(\left(z-z_{a}\right)\left(z-z_{b}\right)\right)\right) \\ &= \frac{1}{1+z^{2}}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+i\operatorname{arg}\left(z-z_{a}\right)+i\operatorname{arg}\left(z-z_{b}\right)\right). \end{align} $$ But for the equalities to hold, define $\displaystyle \operatorname{arg}(z-z_a) \in \left(\frac{3\pi}{4},\frac{11\pi}{4}\right)$ and $\displaystyle \operatorname{arg}(z-z_b) \in \left(-\frac{11\pi}{4},-\frac{3\pi}{4}\right)$. Here is a visual of the keyhole contour.

Keyhole Contour

By Cauchy's Residue Theorem, we get

$$2\pi i \operatorname{Res}(f(z),z=i) = \left(\int_{-R}^{R}+\int_{\Gamma}+\int_{\lambda_1}+\int_{\gamma}+\int_{\lambda_2}\right)f(z)dz.$$ (I forgot to put this in the picture, but $\gamma$ has a small radius $r$ and the gaps between the branch cut and the $\lambda$s have a small length of $\epsilon$).

As $R \to \infty$ and $r \to 0$, it can be proved that $\displaystyle \int_{\Gamma}f(z)dz$ and $\displaystyle \int_{\gamma}f(z)dz$ go to $0$.

For the contour integrals over $\lambda_1$ and $\lambda_2$, let them approach the branch cut $\Lambda$. So $$ \begin{align} & \lim_{R \to \infty}\lim_{\lambda_1, \lambda_2 \to \Lambda}\left(\int_{\lambda_1} + \int_{\lambda_2}\right)f(z)dz \\ =& \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}f\left(z+i\epsilon\right)d\left(z+i\epsilon\right) + \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{z_{a}}^{Rz_{a}}f\left(z-i\epsilon\right)d\left(z-i\epsilon\right) \\ =& \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}\frac{d\left(z+i\epsilon\right)}{1+\left(z+i\epsilon\right)^{2}}\left(\log\left|z+i\epsilon-z_{a}\right|+\log\left|z+i\epsilon-z_{b}\right|+i\operatorname{arg}\left(z+i\epsilon-z_{a}\right)+i\operatorname{arg}\left(z+i\epsilon-z_{b}\right)\right) \\ &+ \lim_{R \to \infty}\lim_{\epsilon \to 0}\int_{Rz_{a}}^{z_{a}}\frac{d\left(z-i\epsilon\right)}{1+\left(z-i\epsilon\right)^{2}}\left(\log\left|z-i\epsilon-z_{a}\right|+\log\left|z-i\epsilon-z_{b}\right|+i\operatorname{arg}\left(z-i\epsilon-z_{a}\right)+i\operatorname{arg}\left(z-i\epsilon-z_{b}\right)\right) \\ =& -\int_{z_a}^{i\infty}\frac{dz}{1+z^2}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+\frac{11\pi i}{4}+i\operatorname{arg}\left(z-z_{b}\right)\right) \\ &+ \int_{z_a}^{i\infty}\frac{dz}{1+z^2}\left(\log\left|z-z_{a}\right|+\log\left|z-z_{b}\right|+\frac{3\pi i}{4}+i\operatorname{arg}\left(z-z_{b}\right)\right) \\ =& -2\pi i \int_{z_a}^{i\infty}\frac{dz}{1+z^2} \\ &= i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right)-\frac{\pi}{2}\ln\left(5\right). \\ \end{align} $$ Next, we will evaluate the residue at the simple pole $z=i$ like this:

$$2\pi i \operatorname{Res}(f(z),z=i) = 2\pi i \lim_{z \to i}\frac{\left(z-i\right)\log\left(\frac{1}{2}+z+z^{2}\right)}{\left(z-i\right)\left(z+i\right)} = \pi\ln\left(\frac{\sqrt{5}}{2}\right)+i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right).$$

Gathering everything together and taking the appropriate limits, we get

$$\pi\ln\left(\frac{\sqrt{5}}{2}\right)+i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right) = \int_{-\infty}^{\infty}f\left(x\right)dx + 0 + i\pi\left(\frac{\pi}{2}+\arctan\left(\frac{1}{2}\right)\right)-\frac{\pi}{2}\ln\left(5\right) + 0.$$

In conclusion,

$$\int_{-\infty}^{\infty}\frac{\ln\left(\frac{1}{2}+x+x^{2}\right)}{1+x^{2}}dx = \pi\ln\left(\frac{5}{2}\right).$$

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    $\begingroup$ Mathematica version 12 is able to evaluate the integral, which does equal $\pi \log \frac{5}{2}$ as claimed. $\endgroup$
    – heropup
    Feb 20, 2023 at 1:37
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    $\begingroup$ A closed form like that screams "Feynman trick". Try $$\frac{1}{2}+x+x^2\to a+\left(x+\frac{1}{2}\right)^2$$ for your parameter. To take advantage of symmetries, you can even shift the entire integrand by $1/4$. $\endgroup$ Feb 20, 2023 at 1:37
  • $\begingroup$ That sounds like a cool idea. I should try that in the future. @NinadMunshi $\endgroup$ Feb 20, 2023 at 3:48
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    $\begingroup$ I made it general in the edit using @Ninad Munshi's suggestion. You can generate nice results $\endgroup$ Feb 20, 2023 at 5:39
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    $\begingroup$ This integral is actually the case $I(1/\sqrt{2}, 1, \frac{\pi}{4}) $ of the integral here. $\endgroup$ Mar 7, 2023 at 1:44

3 Answers 3

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A simpler method using contour integration:

$$\begin{align} \int_{-\infty}^{\infty} \frac{\ln\left(\frac{1}{2}+x+x^{2}\right)}{1+x^{2}} \, \mathrm dx &= 2 \, \Re \int_{-\infty}^{\infty} \frac{\ln \left(x+ \frac{1}{2} + \frac{i}{2} \right)}{1+x^{2}} \, \mathrm dx\\ &= 2 \, \Re \left( 2 \pi i \operatorname{Res} \left[\frac{\ln \left(z+ \frac{1}{2}+\frac{i}{2}\right)}{1+z^{2}}, i \right] \right) \\ &= 2 \, \Re \left(\pi \ln\left( \frac{3i}{2}+ \frac{1}{2} \right)\right) \\ &= 2 \pi \ln \left(\sqrt{\frac{10}{4}} \right) \\ &= \pi \ln \left(\frac{5}{2} \right) \end{align}$$

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  • $\begingroup$ Cool! So I'm assuming this simpler method avoids keyholes by using a semicircular contour in the upper half-plane, correct? @Random Variable $\endgroup$ Feb 20, 2023 at 4:18
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    $\begingroup$ @Accelerator Yes. The branch point is in the lower half-plane. $\endgroup$ Feb 20, 2023 at 4:19
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    $\begingroup$ Short and nice work+1)! $\endgroup$
    – xpaul
    Feb 20, 2023 at 16:27
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Utilize $\int_{0}^{\infty}\frac{\ln\left(a^2+x^{2}\right)}{1+x^{2}}dx = \pi\ln(1+a)$ to integrate \begin{align}\int_{-\infty}^{\infty}\frac{\ln\left(\frac{1}{2}+x+x^{2}\right)}{1+x^{2}}dx =&\int_{0}^{\infty}\frac{\ln\left(\frac{1}{4}+x^{4}\right)}{1+x^{2}}dx= 2\Re \int_{0}^{\infty}\frac{\ln\left(\frac{i}{2}+x^{2}\right)}{1+x^{2}}dx\\ =& \ 2\Re \left(\pi \ln \frac{3+i}2\right)=\pi\ln\frac52 \end{align}

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    $\begingroup$ Good work! I like this. $\endgroup$
    – xpaul
    Feb 20, 2023 at 16:28
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There is even an antiderivative.

Write $$\frac{\log \left(x^2+x+\frac{1}{2}\right)}{x^2+1}=\frac{\log \left((x-a)(x-b)\right)}{(x+i)(x-i)}$$ wtih $$a=-\frac{1+i}{2} \qquad \text{and} \qquad b=-\frac{1-i}{2}$$

$$\frac 1{(x+i)(x-i)}=\frac i 2\left(\frac{1}{x+i}-\frac{1}{x-i}\right)$$ and face four integrals looking like $$\int \frac{\log(x-c)}{x-d}=\text{Li}_2\left(-\frac{x-c}{c-d}\right)+\log (x-c) \log \left(\frac{x-d}{c-d}\right)$$

Edit

As @Ninad Munshi suggested in comments, write $$I(a)=\int_{-\infty}^{+\infty}\frac{\log \left(\left(x+\frac{1}{2}\right)^2+a\right)}{x^2+1}$$ $$I'(a)=\int_{-\infty}^{+\infty}\frac{dx}{\left(x^2+1\right) \left(\left(x+\frac{1}{2}\right)^2+a\right)}$$ $$I'(a)=\frac{4 \pi \left(1+\frac{1}{\sqrt{a}}\right)}{4 a+8 \sqrt{a}+5}$$ $$\int_0^b I(a)\,da=\pi \log \left(\frac{4 b+8 \sqrt{b}+5}{5}\right) $$ and $$I(0)= \pi \log \left(\frac{5}{4}\right)$$

So, as a total $$\int_{-\infty}^{+\infty}\frac{\log \left(\left(x+\frac{1}{2}\right)^2+b\right)}{x^2+1}=\pi \log \left(b+2 \sqrt{b}+\frac{5}{4}\right)$$

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