# How to solve a hard integral?

How prove $\displaystyle \int _{ 0 }^{ \infty }{ (1+x)\arctan { (x) } } \log^4 { (x) }{\frac{1}{\sqrt{x}(1+x^2)}} dx=\frac{57\pi^6\sqrt{2}}{64}$

I found this integral using numerical values.I think the result is correct! How to prove this result?

• A preliminary substitution you could try is $x=\tan t$ but I have no idea if it's the correct approach. Mar 10, 2015 at 23:12
• It might work: replace $x$ with $\tan\theta$ then compute a couple of Fourier series and pray you are able to integrate their product. Mar 10, 2015 at 23:22
• It looks like a tough integral but it is a nice one. Mar 11, 2015 at 19:44

Okay, I'm going to do this the "special functions and pray" way: the first thing to do is produce a simpler integral; the one I'm going to deal with is $$I(a,s) = \int_0^{\infty} \frac{x^{s-1}}{1+x^2} \arctan{\sqrt{a} x} \, dx.$$ (Another option is using $\arctan{(a+x)}$, but that looked even worse when I tried it.)

Now, the required integral is $$\left. \frac{\partial^4}{\partial s^4} (I(1,s)+I(1,s+1)) \right|_{s=1/2} \tag{1};$$ store that away and concentrate on $I$. Now, I don't know about you, but I'm still not convinced I know how to do $I$ directly, so begin with $\partial I/\partial a$, which is $$\frac{\partial I}{\partial a} = \frac{1}{2\sqrt{a}}\int_0^{\infty} \frac{x^{s-1}}{1+x^2} \frac{x}{1+a x^2} \, dx;$$ this one we can do using partial fractions or contour integration, for example, in the usual way, giving $$\frac{\partial I}{\partial a} = \frac{\pi}{4\sqrt{a}} \frac{1-a^{(1-s)/2}}{1-a} \sec{\tfrac{1}{2}\pi s}. \tag{2}$$ At this point we see there's going to be trouble at $a=1$. So now we need to compute $I(1,s)$; it is apparent that $I(0,s)=0$, so we can compute $I(1,s) = \int_0^1 \frac{\partial I(a,s)}{\partial a} \, da$. At this point it would be reasonable to worry: the function we have to integrate is, well, a bit weird. Therefore it is at this point that special functions enter the fray: one definition of the digamma function is: $$\psi(\alpha+1) = -\gamma + \int_0^1 \frac{1-y^{\alpha}}{1-y} \, dy.$$ This isn't quite what we've got, but we can write $$\frac{1-a^{(1-s)/2}}{(1-a)a^{1/2}} = \frac{1-a^{1/2}}{(1-a)a^{1/2}} + \frac{1-a^{-s/2}}{1-a} = \frac{1}{\sqrt{a}(1+\sqrt{a})} + \frac{1-a^{-s/2}}{1-a}.$$ The second of these obviously gives our digamma, and the first integrates to $\log{4}$ by elementary means. Therefore we have made it as far as $$I(1,s) = \frac{\pi}{4} \left( \gamma + \log{4} + \psi\left(1-\tfrac{1}{2}s\right) \right)\sec{\tfrac{1}{2}\pi s},$$ and now we just have to take the four derivatives and evaluate them.

Thankfully at this point another trick suggests itself: $I(1,s)+I(1,2-s)$ has the same $0,2,4,\dotsc$th derivatives at $s=1/2$ as $I(1,s)+I(1,1+s)$. Therefore we can take this sum: \begin{align*} I(1,s)+I(1,2-s) &= \frac{\pi}{4} \left( \gamma + \log{4} + \psi\left(1-\tfrac{1}{2}s\right) \right)\sec{\tfrac{1}{2}\pi s} + \frac{\pi}{4} \left( \gamma + \log{4} + \psi\left(1-\tfrac{1}{2}(2-s)\right) \right)\sec{\tfrac{1}{2}\pi(2- s)} \\ &= \frac{\pi}{4} \left( \psi\left(1-\tfrac{1}{2}s\right) - \psi\left(\tfrac{1}{2}s\right) \right) \sec{\tfrac{1}{2}\pi s} \\ &=\frac{\pi^2}{4} \csc{\tfrac{1}{2}\pi s}, \end{align*} using the reflection formula. It is now straightforward to find fourth derivative at $s=1/2$; it is indeed $$\left. \frac{\partial^4}{\partial s^4} (I(1,s)+I(1,s+1)) \right|_{s=1/2} = \frac{57\pi^6\sqrt{2}}{64}.$$

I didn't actually have to get my digamma on (I just felt like evaluating $I(1,s)$, in the end). Instead, look at (2), and do the same trick with sending $s \mapsto 2-s$. Then the sum becomes $$\frac{\partial}{\partial a} (I(a,s)+I(a,2-s) ) = \frac{\pi}{4\sqrt{a}} \left( \frac{1-a^{(1-s)/2}}{1-a} \sec{\tfrac{1}{2}\pi s} + \frac{1-a^{(s-1)/2}}{1-a} \sec{\tfrac{1}{2}\pi (2-s)} \right) \\ = \frac{\pi}{4} \frac{a^{s/2-1}-a^{-s/2}}{1-a} \sec{\tfrac{1}{2}\pi s}$$ The fraction may look a little familiar to some: integrating from $0$ to $1$, we find $$I(a,s)+I(a,2-s) = \frac{\pi}{4} \sec{\tfrac{1}{2}\pi s} \int_0^1 \frac{a^{s/2-1}-a^{-s/2}}{1-a} \, da.$$ The integral is most easily computed by expanding $(1-a)^{-1}$ as a power series and integrating term-by-term, from which we obtain $$\int_0^1 \frac{a^{s/2-1}-a^{-s/2}}{1-a} \, da = \pi \cot{\tfrac{1}{2}\pi s},$$ (see here for a derivation) (You could also change variables on the second term and consider it as a principal value integral), and then proceed as before. (Remarkably, this makes evaluating the integral all-but elementary: we only need the cotangent series and differentiation under the integral sign).

Edited again:

I think what @Francois Jaclot's answer is getting at is that the function $$\frac{x^{1/2}+1/x^{1/2}}{x+1/x}(\log{x})^4$$ is invariant under the transformation $x \mapsto 1/x$, and $$\arctan{x} + \arctan{\left(\frac{1}{x}\right)} = \frac{\pi}{2}.$$ The integration operator $\int_1^{\infty} \frac{dx}{x}$ maps to $\int_0^1 \frac{dx}{x}$. It follows that, calling the integral we want $I$, $$I = \int_0^{1} \frac{x^{1/2}+1/x^{1/2}}{x+1/x}(\log{x})^4 \arctan{x} \frac{dx}{x} + \int_0^{1} \frac{x^{1/2}+1/x^{1/2}}{x+1/x}(\log{x})^4 \arctan{\left(\frac{1}{x}\right)} \frac{dx}{x} \\ = \frac{\pi}{2} \int_0^{1} \frac{x^{1/2}+x^{-1/2}}{1+x^2}(\log{x})^4 \, \frac{dx}{x}$$ Substituting $y=x^2$, $dy/y = 2dx/x$, so $$I = \frac{\pi}{2 \cdot 2 \cdot 16} \int_0^{1} \frac{y^{1/4}+y^{-1/4}}{1+y} (\log{y})^4 \frac{dy}{y}.$$ Applying the transformation $y \mapsto 1/y$ again, we can rewrite the $y^{-1/4}$ term as an integral on $[1,\infty)$, ending up with: \begin{align*} I &= \frac{\pi}{64} \left( \int_0^{1} \frac{y^{1/4}}{1+y} (\log{y})^4 \frac{dy}{y} + \int_1^{\infty} \frac{y^{1/4}}{1+y} (\log{y})^4 \frac{dy}{y}. \right) \\ &= \frac{\pi}{64} \int_0^{\infty} \frac{y^{-3/4}}{1+y} (\log{y})^4 dy \\ &= \frac{\pi}{64} \left. \frac{d^4}{ds^4}\int_0^{\infty} \frac{y^{s-1}}{1+y} \, dy \right|_{s=1/4} \\ &= \frac{\pi}{64} \left. \frac{d^4}{ds^4} \pi \csc{\pi s} \right|_{s=1/4} \end{align*}, giving the answer again. (I don't like this method as much: all this $1/x$ transforming is quite tricky to get right...)

• This is a fantastic answer +1 Mar 11, 2015 at 3:04
• Chappers is legit +1 Mar 11, 2015 at 5:31
• (+1) "Special functions and pray" is now the name for a wonderful technique. Mar 11, 2015 at 14:43
• Wonderful, very nice demonstration.Thanks(+1) Mar 12, 2015 at 0:29

There is a simpler way, considering $\arctan(x)$ can be easily eliminated. Hence the integral becomes $1/32$th of the fourth derivative of Beta$(t, 1-t)$ at $t = 1/4$.

• Maybe you should expand this answer with some more details? Mar 14, 2015 at 20:14