An arctan integral $\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$

According to Mathematica, we have that

$$\int_0^{\infty } \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx=\pi \left(\frac{\log (2)}{2}-\frac{1321}{6144}\right)$$ that frankly speaking looks pretty nice.

However Mathematica shows that

$$\int \frac{\arctan(x)}{x \left(x^2+1\right)^5} \, dx$$ $$=-\frac{1}{2} i \text{Li}_2\left(e^{2 i \tan ^{-1}(x)}\right)-\frac{1}{2} i \tan ^{-1}(x)^2+\tan ^{-1}(x) \log \left(1-e^{2 i \tan ^{-1}(x)}\right)-\frac{65}{256} \sin \left(2 \tan ^{-1}(x)\right)-\frac{23 \sin \left(4 \tan ^{-1}(x)\right)}{1024}-\frac{5 \sin \left(6 \tan ^{-1}(x)\right)}{2304}-\frac{\sin \left(8 \tan ^{-1}(x)\right)}{8192}+\frac{65}{128} \tan ^{-1}(x) \cos \left(2 \tan ^{-1}(x)\right)+\frac{23}{256} \tan ^{-1}(x) \cos \left(4 \tan ^{-1}(x)\right)+\frac{5}{384} \tan ^{-1}(x) \cos \left(6 \tan ^{-1}(x)\right)+\frac{\tan ^{-1}(x) \cos \left(8 \tan ^{-1}(x)\right)}{1024}$$

and this form doesn't look that nice.

Having given the nice form of the closed form I wonder if we can find a very nice and simple way of getting the answer. What do you think?

A supplementary question:

$$\int_0^{\infty } \frac{\arctan^2(x)}{x \left(x^2+1\right)^5} \, dx=\frac{55}{108}-\frac{1321}{12288}\pi^2+\frac{\pi^2}{4} \log (2)-\frac{7 }{8}\zeta (3)$$

• The "nice and simple way" I can think of is probably going to involve turning this into a complex analysis problem, as the poles are sort of nice and you just need to be careful about branch cuts for the arctan term.. – DaveNine Aug 19 '15 at 18:55
• @DaveNine thank you (+1) Let me know if you also have some ideas about doing this by real methods. – user 1357113 Aug 19 '15 at 19:02
• The supplementary integral is a real bitch... only the dirty by parts method seems to work. but an interesting fact: only the first two parts of the sum seem to be depend on $n=5$ , the second half is already there for $n=1$ – tired Aug 19 '15 at 20:04
• @tired Indeed. I think Umberto P.'s way offers us a nice way also for the second integral. – user 1357113 Aug 19 '15 at 20:10
• Alternatively, since for the first integral, $$\frac{1}{2}\int^\infty_{-\infty}\frac{\arctan{x}}{x(a+x^2)}\ {\rm d}x=-\Re\pi \operatorname*{Res}_{z=i\sqrt{a}}\frac{\ln(1-iz)}{z(z^2+a)}=\frac{\pi\ln(1+\sqrt{a})}{2a}$$ differentiating with respect to $a$ four times gives \begin{align} \int^\infty_0\frac{\arctan{x}}{x(a+x^2)^5}\ {\rm d}x =&\ \frac{\pi\ln(1+\sqrt{a})}{2a^5}-\frac{93\pi}{256a^{9/2}(1+\sqrt{a})}-\frac{29\pi}{256a^{4}(1+\sqrt{a})^2}\\ &-\frac{7\pi}{192a^{7/2}(1+\sqrt{a})^3}-\frac{\pi}{128a^3(1+\sqrt{a})^4} \end{align} Plugging in $a=1$ gives us the desired value. – M.N.C.E. Aug 20 '15 at 12:13

I'm not sure if this is nice and simple, but here is one way:

Substitute $y = \arctan x$ to rewrite as $$\int_0^{\pi/2} \frac{y}{\tan y \sec^{8} y} \, dy$$ Now write this as an integral in $y$ and $\sec y$: $$\int_0^{\pi/2} y \cdot \frac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y} \, dy$$

You can integrate this by parts. Finding the antiderivative of $\dfrac{\sec y \tan y}{(\sec^2 y - 1) \sec^{9} y}$ amounts (after a change of variable) to finding the antiderivative of $$\frac{1}{t^{11} - t^{9}} = \frac{1}{t^{9}(t-1)(t+1)}.$$ This has an elementary partial fractions decomposition which I don't have the energy to carry out.

• I chose your answer for its simplicity, and at the same time for its power that can be successfully used to the supplementary question offering us the possibility to evaluate it pretty easily. – user 1357113 Aug 19 '15 at 20:16
• The substitution $u = t^2$ makes integration slightly less tedious. Alternatively, add and subtract $t^{10}$ from the numerator and split by inspection, giving a simple geometric series and a hyperbolic part. – Jack Lam Nov 7 '16 at 9:39

Here is another solution: Let $(I_n)$ by

$$I_n = \int_{0}^{\infty} \frac{\arctan x}{x(1+x^2)^n} \, dx = \int_{0}^{\frac{\pi}{2}} \frac{\theta}{\sin\theta} \cos^{2n-1}\theta \, d\theta.$$

Then by a simple calculation,

$$I_n - I_{n+1} = \int_{0}^{\frac{\pi}{2}} \theta \sin\theta \cos^{2n-1}\theta \, d\theta = \frac{1}{2n} \int_{0}^{\frac{\pi}{2}} \cos^{2n}\theta \, d\theta.$$

Since $I_n \to 0$ as $n \to \infty$, we find that

$$I_n = \sum_{k=n}^{\infty} \frac{1}{2k} \int_{0}^{\frac{\pi}{2}} \cos^{2k}\theta \, d\theta.$$

Splitting the summation as $\sum_{k=n}^{\infty} = \sum_{k=1}^{\infty} - \sum_{k=1}^{n-1}$, we find that

$$I_n = \frac{\pi}{2}\left( \log 2 - \sum_{k=1}^{n-1} \frac{1}{2k} \frac{(2k-1)!!}{(2k)!!} \right),$$

where $n!!$ denotes the double factorial.

Edit 1. In general, we have

$$\int_{0}^{\infty} \frac{\arctan^s x}{x(1+x^2)^{n+1}} \, dx = \int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta - \sum_{k=1}^{n} \int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta. \tag{1}$$

Currently I have no idea how to obtain a simple formula for the following integral

$$\int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta, \tag{2}$$

even when $s = 2$. On the other hand, for any $s > 0$ and $N \geq \lfloor s/2 \rfloor$ we have

\begin{align*} \int_{0}^{\frac{\pi}{2}} \theta^s \cot \theta \, d\theta &= 2^{-s}\cos\left(\frac{\pi s}{2}\right)\Gamma(1+s)\zeta(1+s) \\ &\quad + \left(\frac{\pi}{2}\right)^s \sum_{k=0}^{N} (-1)^k \pi^{-2k} \frac{\Gamma(2k-s)}{\Gamma(-s)} \eta(2k+1) \\ &\quad + \frac{(-1)^{N+1}}{2^s \Gamma(-s)} \int_{0}^{\infty} \frac{t^{2N+1-s}}{1+t^2} \left( \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n^{1+s}} e^{-\pi n t} \right) \, dt, \end{align*}

where $\eta(s)$ denotes the Dirichlet eta function. (My solution is somewhat involved, so I will post later if it seems useful to our problem.) In particular, when $s$ is a positive integer, then the integral part vanishes and the formula becomes much simpler. Thus the formula (*) gives a closed form as long as we can figure out the integral (2).

Example 1. For example, when $s = 2$ then we can use $N = 1$ and then \begin{align*} \int_{0}^{\frac{\pi}{2}} \theta^2 \cot \theta \, d\theta &= -\frac{1}{2}\zeta(3) + \frac{\pi^2}{4} \log 2 - \frac{1}{2}\eta(3) \\ &= \frac{\pi^2}{4}\log 2 - \frac{7}{8}\zeta(3). \end{align*} Since we can figure out the integral (2) for $s = 2$ and $k = 1, \cdots, 4$, we easily obtain OP's last identity.

Here is another example:

Example 2. Using the formula with $s = 6$, we can check that \begin{align*} \int_{0}^{\infty} \frac{\arctan^6 x}{x(1+x^2)^3} \, dx &= \frac{\pi^6}{64} \log 2 -\frac{45 \pi^4}{128} \zeta(3) + \frac{675 \pi^2}{128} \zeta(5) -\frac{5715}{256} \zeta(7) \\ &\quad - \frac{11 \pi^6}{2048} + \frac{705 \pi^4}{4096} - \frac{8595 \pi^2}{4096} + \frac{135}{16} \\ &\approx 0.0349464822054751922142122595622\cdots. \end{align*}

• Very nice. I was waiting a solution for (one of) the general case(s). (+1) :-) – user 1357113 Aug 21 '15 at 6:38
• Right, well done! I like $\int_{0}^{\frac{\pi}{2}} \theta^s \sin\theta \cos^{2k-1}\theta \, d\theta$ but first I'd prefer to try it with positive integer powers. – user 1357113 Aug 21 '15 at 8:26
• For the case $s=2$ I think we can start with the integration by parts, and then all reduces to an integral that possibly is already calculated in the Irresistible Integrals book. I will check that now. – user 1357113 Aug 21 '15 at 8:34
• If you have the foregoing book, check the problem 11.3.9, page 228. – user 1357113 Aug 21 '15 at 8:46

What about Feynman's way? If we take: $$f(a) = \int_{0}^{+\infty}\frac{\arctan(ax)}{x(1+x^2)^5}\,dx$$ we have $f(0)=0$ and: $$f'(a)=\int_{0}^{+\infty}\frac{dx}{(1+x^2)^5 (1+a^2 x^2)}$$ that is a manageable integral through partial fraction decomposition / the residue theorem.

We have: $$\begin{eqnarray*} f'(a) &=& \frac{\pi}{256(1+a)^5}\left(35+175 a+345 a^2+325 a^3+128 a^4\right)\\&=&\frac{\pi}{256(1+a)^5}\left(8-52(a+1)+138 (a+1)^2-187(a+1)^3+128(a+1)^4\right)\end{eqnarray*}$$ and it is not difficult to integrate such expression over $(0,1)$.

• Indeed (+1) So many ways out there. – user 1357113 Aug 19 '15 at 19:42
• Feynmann lives! Jack, I was writing a solution using Differentiation Under the Integral Sign when I noticed your post. Foiled again! ;-)) +1 to you. I have to ask you, did you use a computer tool to do the dirty work of cleaning up the form for $f'$? – Mark Viola Aug 19 '15 at 19:52
• @Dr.MV: yes I did. I was too afraid to compute a wrong coefficient. Anyway, to compute $f'$ by hand is not so terrible, just lengthy. – Jack D'Aurizio Aug 29 '15 at 13:37
• Yes, a bit tedious. – Mark Viola Aug 29 '15 at 15:13

Does it look any nicer? $$\int\frac{\arctan x}{x\,\left(x^2+1\right)^5}\, dx=\frac12\,\Im\operatorname{Li}_2\left(e^{2\,i\arctan x}\right)\\-\frac x{9216\,\left(x^2+1\right)^4}\left(3963x^6+12995x^4+14525x^2+5637\right)\\+\left[\ln\frac{2x}{\sqrt{x^2+1}}+\frac{12x^6+42x^4+52x^2+25}{24\,\left(x^2+1\right)^4}-\frac{1321}{3072}\right]\cdot\arctan x$$

• (+1) Not that nice ;) – user 1357113 Aug 22 '15 at 10:58