Prove that $\int_{0}^{\infty}\frac{(\arctan x)^3}{x^3}dx=\frac{3π}{2}\ln2-\frac{π^3}{16}$

Let's integrate by parts, setting $ u':x\mapsto\frac{1}{x^{3}} $, and $ v:x\mapsto\arctan^{3}{x} $, we get : \begin{aligned}\int_{0}^{+\infty}{\frac{\arctan^{3}{x}}{x^{3}}\,\mathrm{d}x}&=\left[-\frac{\arctan^{3}{x}}{2x^{2}}\right]_{0}^{+\infty}+\frac{3}{2}\int_{0}^{+\infty}{\frac{\arctan^{2}{x}}{x^{2}\left(1+x^{2}\right)}\,\mathrm{d}x}\\ &=\frac{3}{2}\int_{0}^{+\infty}{\arctan^{2}{x}\left(\frac{1}{x^{2}}-\frac{1}{1+x^{2}}\right)\mathrm{d}x}\\ &=\frac{3}{2}\int_{0}^{+\infty}{\left(\frac{\arctan{x}}{x}\right)^{2}\,\mathrm{d}x}-\frac{1}{2}\int_{0}^{+\infty}{\frac{3\arctan^{2}{x}}{1+x^{2}}\,\mathrm{d}x}\\ &=\frac{3}{2}\pi\ln{2}-\frac{1}{2}\left[\arctan^{3}{x}\right]_{0}^{+\infty}\\ &=\frac{3\pi}{2}\ln{2}-\frac{\pi^{3}}{16}\end{aligned}

I want to illustrate an alternative way of attacking this integral using Parseval’s Theorem.

We begin by expressing

$$\frac{\arctan{x}}{x} = \int_0^1 \frac{dt}{1+x^2 t^2} $$

Then the integral is equal to, after changing to order of integration,

$$\int_0^1 dt \, \int_0^1 du \, \int_0^1 dv \, \int_0^{\infty} \frac{dx}{(1+x^2 t^2) (1+x^2 u^2) (1+x^2 v^2)} $$

It is understandable that this approach so far does not look promising. Nevertheless, as an alternative approach it is worth exploring. For example, using Parseval’s theorem on the inner integral looks tricky, as the integrand is a product of three rather than two functions. In this case, however, we can consider a product of two of the functions to be one function so we can apply Parseval’s theorem. By the convolution theorem, we find that

$$\int_{-\infty}^{\infty} dx \, f(x) g(x) h(x) = \frac1{4 \pi^2} \int_{-\infty}^{\infty} dk’ F(k’) \int_{-\infty}^{\infty} dk \, G(k-k’) H(k) $$

In this case, $f(x) = 1/(1+x^2 t^2)$, etc. Accordingly, $F(k) = (\pi/t) e^{-|k|/t}$, $G(k) = (\pi/u) e^{-|k|/u}$, $H(k) = (\pi/v) e^{-|k|/v}$. Plugging this into the RHS of the above Parseval relation, we get for the integral

$$\frac{\pi}{4 t u v} \int_{-\infty}^{\infty} dk’ \, e^{-|k’|/t} \, \int_{-\infty}^{\infty} dk \, e^{-|k-k’|/u} e^{-|k|/v} $$

Note that the above integral, while elementary, is a challenge for a CAS like Mathematica. (I have yet to see Mathematica evaluate this integral without any further simplification.) Note that the above integral is twice the result we seek. We can evaluate the integral by restricting $k’ \ge 0$ as follows:

$$\frac{\pi}{4 t u v} \int_0^{\infty} dk’ \, e^{-k’/t} \, \int_{-\infty}^{\infty} dk \, \left ( e^{-|k-k’|/u} + e^{-|k+k’|/u} \right ) e^{-|k|/v} $$

The inner integral may be evaluated as follows.

$$ \begin{align} \int_{-\infty}^{\infty} dk \, \left ( e^{-|k-k’|/u} + e^{-|k+k’|/u} \right ) e^{-|k|/v} &= e^{-k’/u} \left [\int_{-\infty}^0 dk \, e^{\left ( \frac1{u}+\frac1{v} \right ) k} + \int_0^{k’} dk \, e^{\left ( \frac1{u}-\frac1{v} \right ) k} \right ] \\ &+ e^{k’/u} \int_{k’}^{\infty} dk \, e^{-\left ( \frac1{u}+\frac1{v} \right ) k} \\ &+ e^{k’/u} \int_{-\infty}^{-k’} dk \, e^{\left ( \frac1{u}+\frac1{v} \right ) k} \\ &+ e^{-k’/u} \left [ \int_{-k’}^0 dk \, e^{-\left ( \frac1{u}-\frac1{v} \right ) k} + \int_0^{\infty} dk \, e^{- \left ( \frac1{u}+\frac1{v} \right )k} \right ] \end{align} $$

Now that all of the absolute values have been stripped from the exponential, these integrals are all elementary. The inner integral is, after simplification,

$$4 u v \frac{u e^{-k’/u} - v e^{-k’/v}}{u^2-v^2} $$

The integral we seek is then

$$\frac{\pi}{t (u^2-v^2)} \int_0^{\infty} dk’ \, e^{-k’/t} \left ( u e^{-k’/u} - v e^{-k’/v} \right ) $$

which evaluates to

$$\pi \frac{u v + u t + v t}{(u+v)(u+t)(v+t)} $$

While this is a nice, simple expression for that terrible-looking integral, we still have to integrate this expression over $(t,u,v) \in [0,1]^3$. I found that we can make things easier on ourselves with some algebraic manipulation. I leave it to the reader to show that the triple integral may be transformed algebraically to

$$\frac{\pi}{2} \int_0^1 dt \, \int_0^1 du \, \int_0^1 dv \, \left [ \frac{u}{(u+t)(u+v)}+\frac{v}{(v+u)(v+t)} + \frac{t}{(t+u)(t+v)} \right ] $$

which by symmetry is simplified to

$$\frac{3 \pi}{2} \int_0^1 dt \, \int_0^1 du \, \int_0^1 dv \, \frac{t}{(t+u)(t+v)} $$

The inner two integrations are trivial and we are left with a single integral:

$$\frac{3 \pi}{2} \int_0^1 dt \, t \log^2{\left (1+\frac1{t} \right )} $$

Let’s pause to assess. We began with a tough-looking integral and we are now left with another tough integral. It is uncertain what progress we have made after all of the manipulations we performed above. But upon further reflection, the transformed integral is one that may be attacked directly with relatively elementary methods. For example, by expanding the log squared term and combining like terms, we get the following expression for the above integral:

$$\frac{3 \pi}{2} \left (\int_0^2 dt \, t \, \log^2{t} - \int_1^2 dt \, \log^2{t} \right ) - 3 \pi \int_0^1 dt \, t \log{(1+t)} \log{t} $$

The integrals inside the parentheses combine to be equal to $2 \log{2}-1$. The final integral is evaluated by Taylor expanding the $\log{(1+t)}$ term in the integrand; as a result we get a sum:

$$3 \pi \sum_{k=1}^{\infty} \frac{(-1)^k}{k (k+2)^2} $$

By partial fractions we can deduce that

$$\frac1{k (k+2)^2} = \frac14 \frac1{k} - \frac14 \frac1{k+2} - \frac12 \frac1{(k+2)^2} $$

Based on this and the facts that

$$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} = \log{2}$$ $$\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2} = \frac{\pi^2}{12} $$

we get that

$$ \sum_{k=1}^{\infty} \frac{(-1)^k}{k (k+2)^2} = \frac{\pi^2}{24} - \frac12$$

And finally we can combine all of the above and divide by $2$ (because the integral is over $[0,\infty)$) to get the result we seek:

$$\begin{align} \int_0^{\infty} dx \, \left ( \frac{\arctan{x}}{x} \right )^3 &= \frac12 \left [\frac{3 \pi}{2} (2 \log{2}-1) - 3 \pi \left ( \frac{\pi^2}{24} - \frac12 \right ) \right ] \\ &= \frac{3 \pi}{2} \log{2} - \frac{\pi^3}{16} \end{align}$$