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While browsing similar questions on this site I came up with the following integral because I thought I could evaluate it. $$I=\int_{-\infty}^{\infty}\frac{x\arctan x\ \log(1+x^2)}{1+x^2}dx$$ I've been able to simplify it a bit. We first notice that $$I=\int_0^{\infty}\frac{\arctan x\ \log(1+x^2)}{1+x^2}2xdx$$ The substitution $u=x^2+1$ gives $$I=\int_1^{\infty}\arctan\sqrt{u-1}\ \log u\ \frac{du}u$$ Then $w=\log u$ gives $$I=\int_{0}^{\infty}\arctan\sqrt{e^w-1}\ dw$$ Which I do not know how to proceed with.

Another approach I tried was this. Starting with the original integral,

$x=\tan u$: $$I=2\int_0^{\pi/2}u\tan u\log\sec^2u\ du$$ Which I also do not know how to do. Please help me proceed or give me a value of the integral (and show how you got it).

If no closed form exists (AKA you have an answer in terms of a series or special function), I'm fine with that.

cheers!

Edit: In the comments it is discussed that the integral is not integrable over the positive reals, but the following related integral is:

$$J=\int_0^{\infty}\frac{\log(1+x^2)\arctan\frac1x}{1+x^2}xdx$$ So. How do we find the value for $J$?

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  • $\begingroup$ No, wait, as $x\to +\infty$ the integrand function behaves like $\frac{\pi x \log(x)}{1+x^2}$, which is not integrable on $\mathbb{R}^+$. The strategy in my deleted answer (integration by parts and second derivative of a Beta function) works if $\arctan(x)$ is replaced by $\arctan(1/x)$. $\endgroup$ Nov 4, 2018 at 3:52
  • $\begingroup$ It leads to $$\int_{\mathbb{R}}\frac{x\arctan(1/x)\log(1+x^2)}{1+x^2}\,dx = \frac{\pi^3}{12}+\pi\log^2(2).$$ $\endgroup$ Nov 4, 2018 at 3:54
  • $\begingroup$ @JackD'Aurizio so the original integral cannot be found? $\endgroup$
    – clathratus
    Nov 4, 2018 at 3:56
  • $\begingroup$ The original integral is simply divergent. But I guess you are still in time to update the post and replace $\arctan(x)$ with $\arctan(1/x)$. $\endgroup$ Nov 4, 2018 at 3:57
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    $\begingroup$ I am sure, no matter what a CAS can say, $\frac{x}{1+x^2}$ is not integrable on $\mathbb{R}^+$. (And you are just pointing out that $f(x)\to 0$ as $x\to +\infty$, which is not a sufficient condition for integrability) $\endgroup$ Nov 4, 2018 at 4:00

6 Answers 6

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Integration by parts reduces the original problem to the evaluation of $$ \int_{0}^{+\infty}\frac{\log^2(1+x^2)}{1+x^2}\,dx $$ which is pretty straightforward: since $$ \int_{0}^{+\infty}(1+x^2)^{s-1}\,dx =\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{1}{2}-s\right)}{\Gamma(1-s)}$$ by applying $\frac{d^2}{ds^2}$ to both sides, then considering $\lim_{s\to 0^+}$, we get: $$ \int_{0}^{+\infty}\frac{\log^2(1+x^2)}{1+x^2}\,dx =\frac{\pi^3}{6}+2\pi\log^2(2).$$ You may find another example of this technique at page 81 of my notes.

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  • $\begingroup$ I'm a bit lost on your train of thinking, could you elaborate on your steps? $\endgroup$
    – clathratus
    Nov 4, 2018 at 3:49
  • $\begingroup$ Okay I looked on page 81 of your notes and I am still confused. In your proof of $$K=\int_0^{\pi/2}\log^3(\sin x)dx=-\frac{\pi}{8}(\pi^2\log2+4\log^3 2+6\zeta(2))$$ You say that differentiation under the integral sign gives $$K=.5\Gamma(.5)D^3\frac{\Gamma(\frac{\alpha+1}2)}{\Gamma(1+\frac\alpha2)}\bigg|_{\alpha=0}$$ Where in the integral did you put $\alpha$? $\endgroup$
    – clathratus
    Nov 4, 2018 at 19:02
  • $\begingroup$ @clathratus: $$\log(\sin x) = \left.\frac{d}{d\alpha}(\sin x)^{\alpha}\right|_{\alpha=0}$$ $\endgroup$ Nov 4, 2018 at 19:03
  • $\begingroup$ Oh. wow I feel like an idiot. Thanks! $\endgroup$
    – clathratus
    Nov 4, 2018 at 19:04
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\begin{align*} I(a,b)&=\int_{0}^{+\infty}\frac{x\ln\left(1+b^{2}x^{2}\right)\arctan\left(bx\right)}{a^{4}+x^{4}}\,\mathrm{d}x\\ &=\Im\left[\int_{0}^{+\infty}\frac{z\ln^2(1+ibx)}{a^{4}+x^{4}}\,\mathrm{d}x\right] \end{align*}

Now the residue theorem give

$$\color{red}{\frac{\pi}{2a^{2}}\ln\left(1+\sqrt{2}ab+a^{2}b^{2}\right)\arctan\left(\frac{ab}{\sqrt{2}+ab}\right)\,}$$

How about below integral?

$$\int_{0}^{+\infty}\frac{x\ln\left(1+b^{2}x^{2}\right)\arctan\left(cx\right)}{a^{4}+x^{4}}\,\mathrm{d}x$$

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Let $g(x)$ be $2x\ln(1 + x^{2})$

Let $f(x)$ be $\arctan(x)$

$2I = \int g(x)f(x)f'(x)dx$

Note that:

  • $f'(x) = \frac{1}{1+x^{2}}$
  • Let $u = f(x)$. Then, $F(x) = \int f(x)f'(x)dx = \int u du = \frac{1}{2}f(x)^{2}$
  • Let $v = 1 + x^{2}$. Then, $G(x) = \int \ln(v)dv = v[\ln(v) - 1] = (1 + x^{2})[\ln(1 + x^{2}) - 1] $

$\color{red}{2I = f(x)G(x) - \int f'(x)G(x)dx}$

$ = f(x)G(x) - \int (1 + x^{2})[\ln(1 + x^{2}) - 1] \frac{1}{1+x^{2}} dx$

$ = f(x)G(x) - \int \ln(1+x^{2}) - 1 dx$

$ = f(x)G(x) + x - \int \ln(1+x^{2}) dx$

Let $a(x) = x$ and $b(x) = \ln(1+x^{2})$

$ \int 1 \cdot \ln(1+x^{2}) dx $ $ = \int b(x) a'(x) dx = a(x)b(x) - \int a(x)b'(x) dx$ $ \int a(x)b'(x) dx = \int x \frac{2x}{1 + x^{2}} dx $ $ = 2\int \frac{x^{2}}{1 + x^{2}} dx $ $ = 2(x - \arctan(x)) $

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  • $\begingroup$ Wonderful!!! Thank you so much! $\endgroup$
    – clathratus
    Nov 4, 2018 at 17:09
  • $\begingroup$ This is indefinite integral. I think what Jack say about your original problem is correct. $\endgroup$
    – R zu
    Nov 4, 2018 at 17:11
  • $\begingroup$ Would you prefer it if I accepted his answer? $\endgroup$
    – clathratus
    Nov 4, 2018 at 18:49
  • $\begingroup$ Either way is fine. I up-voted his answer because it looks like a useful technique for me. $\endgroup$
    – R zu
    Nov 4, 2018 at 18:51
  • $\begingroup$ @Rzu, you wrote that $$2I = f(x)G(x) - \int f'(x)G(x)dx\;\;,$$ but you should have written that $$2I = f(x)f’(x)G(x) - \int \big(f(x)f'(x)\big)’G(x)dx\;\;,$$ so you did not calculate the integral correctly. $\endgroup$
    – Angelo
    Jan 2, 2023 at 12:40
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Continuing from Jack D'A.'s post-IBP integral:

$$\begin{align*} J &= \int_0^\infty \frac{\log^2(1+x^2)}{1+x^2} \, dx \\[1ex] &= \int_0^1 \frac{\log^2(1+x^2) + \log^2\left(1+\frac1{x^2}\right)}{1+x^2} \, dx \tag{1} \\[1ex] &= \frac12 \int_0^1 \left(\log^2\left(\frac2{1+\sqrt{1-x^2}}\right) + \log^2\left(\frac2{1-\sqrt{1-x^2}}\right)\right) \, \frac{dx}{\sqrt{1-x^2}} \tag{2} \\[1ex] &= \log^2(2) \int_0^1 \frac{dx}{\sqrt{1-x^2}} - \log(2) \int_0^1 \frac{\log(1-x^2)}{\sqrt{1-x^2}}\,dx \\ &\qquad + \frac12 \int_0^1 \frac{\log^2(1+x) + \log^2(1-x)}{\sqrt{1-x^2}} \, dx \tag{3} \\[1ex] &= \log^2(2) \int_0^1 \frac{dx}{\sqrt{1-x^2}} - \log(2) \int_0^1 \frac{\log(1-x^2)}{\sqrt{1-x^2}}\,dx \\ &\qquad + \frac12 \int_0^1 \frac{\log^2(1-x^2)}{\sqrt{1-x^2}} \, dx - \int_0^1 \frac{\log(1+x)\log(1-x)}{\sqrt{1-x^2}} \, dx \tag{4} \\[1ex] &= \log^2(2) \int_0^{\frac\pi2} dx - 2\log(2) \int_0^{\frac\pi2} \log(\cos(x)) \, dx \\ &\qquad + 2 \int_0^{\frac\pi2} \log^2(\cos(x)) \, dx - \underbrace{\int_0^{\frac\pi2} \log(1+\sin(x)) \log(1-\sin(x)) \, dx}_{K} \tag{5} \\[1ex] &= \boxed{2\pi\log^2(2) + \frac{\pi^3}6} \tag{6a,b,c,d} \end{align*}$$


Steps
  • $(1)$ : fold up the integral at $x=1$, i.e. substitute $x\mapsto\dfrac1x$ over $[1,\infty)$
  • $(2)$ : substitute $x\mapsto\dfrac{2x}{x^2+1}$
  • $(3)$ : substitute $x\mapsto\sqrt{1-x^2}$
  • $(4)$ : rewrite the logarithm
  • $(5)$ : substitute $x\mapsto\sin^{-1}(x)$
  • $(6\rm a)$ : the first integral is trivial
  • $(6\rm b)$ : see here for the second integral
  • $(6\rm c)$ : see here for the third integral
  • $(6\rm d)$ : substitute $x\mapsto\dfrac\pi2-x$ and follow J.D'A.'s comment here to evaluate $K$ by expanding into Fourier series:

$$\begin{align*} K &= \int_0^{\frac\pi2} \log(1+\cos(x)) \log(1-\cos(x)) \, dx \\[1ex] &= \int_0^{\frac\pi2} \left(\log(2) + \sum_{k=1}^\infty \frac{2\cos(kx)}k\right) \left(\log(2) + \sum_{k=1}^\infty \frac{2(-1)^k\cos(kx)}k\right) \, dx \\[1ex] &= \int_0^{\frac\pi2} \left[\log^2(2) + \log(2) \sum_{k=1}^\infty \frac{2\left(1+(-1)^k\right) \cos(kx)}k \right. \\ & \qquad \qquad \left. {} + 4 \left(\sum_{k=1}^\infty \frac{\cos(kx)}k\right) \left(\sum_{k=1}^\infty \frac{(-1)^k\cos(kx)}k\right)\right] \, dx \\[1ex] &= \frac\pi2 \log^2(2) + \log(2) \sum_{k=1}^\infty \frac2k \underbrace{\int_0^{\frac\pi2} \cos(2kx) \, dx}_{=0} \\ &\qquad {} + 4 \left[\sum_{k=1}^\infty \frac{(-1)^k}{k^2} \underbrace{\int_0^{\frac\pi2} \cos^2(kx) \, dx}_{=\pi/4} + 2 \sum_{k=2}^\infty \sum_{\ell=1}^{k-1} \frac{(-1)^k}{k\ell} \int_0^{\frac\pi2} \cos(kx) \cos(\ell x) \, dx\right] \\[1ex] &= \frac\pi2 \log^2(2) + \pi \sum_{k=1}^\infty \frac{(-1)^k}{k^2} \\ &\qquad {} + 4 \sum_{k=2}^\infty \sum_{\ell=1}^{k-1} \frac{(-1)^k}{k\ell} \underbrace{\int_0^{\frac\pi2} \bigg(\cos((k-\ell)x) + \cos((k+\ell)x)\bigg) \, dx}_{=0} \\[1ex] &= \frac\pi2 \log^2(2) - \frac{\pi^3}{12} \end{align*}$$

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We may also try attacking the third integral in line $(3)$ in my first answer,

$$L = \int_0^1 \frac{\log^2(1+x)+\log^2(1-x)}{\sqrt{1-x^2}} \, dx$$

by exploiting the generating function of $\dfrac{H_{2k-1}}k$, with $H_k$ the $k^{\rm th}$ harmonic number. Using the Cauchy product, we find

$$-\log(1\pm x) = \sum_{n=1}^\infty \frac{(\pm x)^n}n \\ \implies \log^2(1\pm x) = \sum_{n=2}^\infty \sum_{m=1}^{n-1} \frac{x^n}{m(n-m)} = \sum_{n=2}^\infty \frac{2 H_{n-1}}n (\pm x)^n \\ \implies \log^2(1+x) + \log^2(1-x) = \sum_{n=2}^\infty \frac{2H_{n-1}}{n} \left(1+(-1)^n\right) (\pm x)^n = \sum_{n=1}^\infty \frac{H_{2n-1}}{n} x^{2n}$$

Multiply by $\frac1{\sqrt{1-x^2}}$ and integrate to recover an Euler sum:

$$\begin{align*} L &= \sum_{n=1}^\infty \frac{H_{2n-1}}{n} \int_0^1 \frac{x^{2n}}{\sqrt{1-x^2}} \, dx \\[1ex] &= \frac12 \sum_{n=1}^\infty \frac{H_{2n-1}}{n} \int_0^1 x^{n-\frac12} (1-x)^{-\frac12} \, dx \\[1ex] &= \frac12 \sum_{n=1}^\infty \frac{H_{2n-1}}{n} \operatorname{B}\left(n+\frac12,\frac12\right) \\[1ex] &= \frac12 \sum_{n=1}^\infty \frac{H_{2n-1}}{n\cdot4^n} \binom{2n}n \end{align*}$$

where in the second line, we substitute $x\mapsto\sqrt x$. Now we can use

$$\sum_{n=1}^\infty \frac{H_n}{n\cdot4^n} \binom{2n}n = \frac{\pi^2}3$$

(see the proof following equation $(20)$) together with

$$H_{2n-1} = \frac12 \left(H_n + H_{n-\frac12}\right) + \log(2)$$

to determine $\displaystyle L=\frac{\pi^3}3+\pi\log^2(2)$.

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Integration by parts yields $$ \begin{aligned} I & =\int_{-\infty}^{\infty} \frac{x \arctan \frac{1}{x} \ln \left(1+x^2\right)}{1+x^2} d x \\ & =2 \int_0^{\infty} \frac{x \arctan \frac{1}{2} \ln \left(1+x^2\right)}{1+x^2} d x \\ & =\frac{1}{2} \int_0^{\infty} \arctan \frac{1}{x} d\left(\ln ^2\left(1+x^2\right)\right) \\ & =\frac{1}{2} \int_0^{\infty} \ln ^2\left(1+x^2\right) \frac{1}{1+\frac{1}{x^2}} \cdot \frac{d x}{x^2} \\ & =\frac{1}{2} \int_0^{\infty} \frac{\ln ^2\left(1+x^2\right)}{1+x^2} d x\end{aligned} $$ Letting $x\mapsto \tan \theta$ transforms the integral into a Beta function. $$ \begin{aligned} I & =2 \int_0^{\frac{\pi}{2}} \ln ^2(\cos \theta) d \theta \\ & =\left.2 \frac{d^2}{d a^2} \int_0^{\frac{\pi}{2}} \cos ^a \theta d \theta\right|_{a=0} \\ & =\left.\frac{d^2}{d a^2} B\left(\frac{1}{2}, \frac{a+1}{2}\right)\right|_{a=0} \\ & =\boxed{\frac{\pi^3}{12}+\pi \ln ^2 2} \end{aligned} $$

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