# How to evaluate $\int_{-\infty}^{\infty}\frac{x\arctan\frac1x\ \log(1+x^2)}{1+x^2}dx$

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.

$$x=\tan u$$: $$I=2\int_0^{\pi/2}u\tan u\log\sec^2u\ du$$ 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 $$J$$?

• 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 $\arctan(1/x)$. Nov 4 '18 at 3:52
• It leads to $$\int_{\mathbb{R}}\frac{x\arctan(1/x)\log(1+x^2)}{1+x^2}\,dx = \frac{\pi^3}{12}+\pi\log^2(2).$$ Nov 4 '18 at 3:54
• @JackD'Aurizio so the original integral cannot be found? Nov 4 '18 at 3:56
• The original integral is simply divergent. But I guess you are still in time to update the post and replace $\arctan(x)$ with $\arctan(1/x)$. Nov 4 '18 at 3:57
• @JackD'Aurizio are you sure? desmos.com/calculator/e2efned0cl Nov 4 '18 at 3:59

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.

• I'm a bit lost on your train of thinking, could you elaborate on your steps? Nov 4 '18 at 3:49
• 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 $\alpha$? Nov 4 '18 at 19:02
• @clathratus: $$\log(\sin x) = \left.\frac{d}{d\alpha}(\sin x)^{\alpha}\right|_{\alpha=0}$$ Nov 4 '18 at 19:03
• Oh. wow I feel like an idiot. Thanks! Nov 4 '18 at 19:04

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, $$G(x) = \int \ln(v)dv = v[\ln(v) - 1] = (1 + x^{2})[\ln(1 + x^{2}) - 1]$$

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

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

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

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

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

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

• Wonderful!!! Thank you so much! Nov 4 '18 at 17:09
• This is indefinite integral. I think what Jack say about your original problem is correct.
– R zu
Nov 4 '18 at 17:11
• Would you prefer it if I accepted his answer? Nov 4 '18 at 18:49
• Either way is fine. I up-voted his answer because it looks like a useful technique for me.
– R zu
Nov 4 '18 at 18:51

\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

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

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