$$I(a)=\int_{0}^{\infty}\sin^3(x)\cos[a\tan(x)]\frac{dx}{x}$$ I'd like to evaluate the integral by differentiating with respect to parameter $a$ but no success yet. Seems impossible. What would the other options?
Edit: A hypothetical closed form solution:
$$I(a)=\frac{\pi}{4}(1-a)e^{-a}$$
Assume $a > 0$ and notice:
We have: $$\begin{align} I(a) = & \int_0^{\infty}\sin^3(x)\cos(a\tan x)\frac{dx}{x}\\ = & \frac12 \int_{-\infty}^{\infty}\sin^3(x)\cos(a\tan x)\frac{dx}{x}\\ = & \frac12 \sum_{n=-\infty}^{\infty} \int_{(n-\frac12)\pi}^{(n+\frac12)\pi}\sin^3(x)\cos(a\tan x)\frac{dx}{x}\\ \overset{(\text{ by }*1)}{=} &\frac12 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^3(x)\cos(a\tan x)\left(\sum_{n=-\infty}^{n=\infty} \frac{(-1)^n}{x + n\pi}\right)dx\\ \overset{(\text{ by }*2)}{=} &\frac12 \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\sin^2(x)\cos(a\tan x)dx \end{align}$$ Now let $t = \tan x$, we get: $$I(a) = \frac12 \int_{-\infty}^{\infty} \frac{t^2}{1+t^2} \cos(at) \frac{dt}{1+t^2} = \frac12 \int_{-\infty}^{\infty} \frac{t^2}{(1+t^2)^2} e^{iat} dt\tag{*3} $$ Decompose $\frac{t^2}{(1+t^2)^2}$ into partial fractions:
$$\frac{t^2}{(1+t^2)^2} = \frac14 \left( \frac{1}{(t-i)^2} - \frac{i}{t-i} + \frac{i}{t+i} + \frac{1}{(t+i)^2}\right)$$ We can integrate $(*3)$ using contour integral. Since $a > 0$, we can deform the contour in upper half plane, pickup residues of the integrand at $t = i$ and get: $$I(a) = \frac{2\pi i}{8}\left[ \frac{d}{dt}\left(e^{iat}\right) - i e^{iat} \right]_{t=i} = \frac{\pi}{4}(1-a) e^{-a} $$
A special case is easy, using $4 \sin^3(x) = 3 \sin(x) - \sin(3x)$: $$ I(0) = \int_0^\infty \frac{\sin^3(x)}{x} \mathrm{d} x = \frac{3}{4} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x - \frac{1}{4} \int_0^\infty \frac{\sin(3x)}{x} \mathrm{d}x = \frac{1}{2} \int_0^\infty \frac{\sin(x)}{x} \mathrm{d} x = \frac{\pi}{4} $$ It is easy to see that $I^\prime(a)$ is singular at $a=0$.
In order to evaluate $I(a)$ numerically some massaging is needed: $$ \begin{eqnarray} I(a) &=& \int_0^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \int_{-\infty}^\infty \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \sum_{n=-\infty}^\infty \int_{-\frac{\pi}{2} + \pi n }^{\frac{\pi}{2} + \pi n} \sin^3(x) \cos(a \tan x) \frac{\mathrm{d}x}{x} \\ &=& \frac{1}{2} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \sum_{n=-\infty}^\infty (-1)^n \sin^3(y) \cos(a \tan y) \frac{\mathrm{d}y}{y+\pi n} \\ &\stackrel{t =\tan(u)}{=} & \frac{1}{2} \int_{-\infty}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u = \int_{0}^\infty \frac{u^3}{(1+u^2)^{5/2}} \cos(a u) f(u) \mathrm{d} u \end{eqnarray} $$ where $f(u)$ is defined by the sum: $$ \begin{eqnarray} f(u) &=& \sum_{n=-\infty}^\infty \frac{(-1)^n}{\pi n + \arctan(u)} = \frac{1}{\arctan(u)} - 2 \arctan(u) \sum_{n=1}^\infty \frac{(-1)^n}{\pi^2 n^2 - \arctan(u)^2} \\ &=& \frac{1}{\pi} \left( \Phi \left(-1,1,\frac{\arctan(u)}{\pi }\right)+\Phi \left(-1,1,1-\frac{\arctan(u)}{\pi }\right) \right) \end{eqnarray} $$ where $\Phi(z,s,a)$ denotes the Lerch transcendent. With this we can plot $I(a)$ using numerical methods:
Let us assume that $f(z)$ has a Maclaurin series expansion with real coefficients that converges on the unit circle on the complex plane.
Then
$$ \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx = \int_{0}^{\infty} \frac{\sin^3 x}{x} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cos(2nx) \, dx .$$
Now assuming we can switch the order of summation and integration (which would be more difficult to justify if the series doesn't converge absolutely),
$$ \begin{align} &\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx \\ &= f(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, dx + f'(0) \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2x) \, dx + \sum_{n=2}^{\infty}\frac{f^{(n)}(0)}{n!} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(2nx) \, dx. \end{align}$$
I will now show that all the integrals vanish except for the first two.
Notice that for $a \ge 0$,
$$ \begin{align} &\int_{0}^{\infty} \frac{\sin^{3} x}{x} \cos(ax) \, dx \\ &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin(x) - \sin (3x)}{x} \cos(ax) \, dx \\ &= \frac{1}{8} \int_{0}^{\infty} \frac{3 \sin [(1+a)x] + 3 \sin [(1-a)x] - \sin [(3+a)x] - \sin [(3-a)x]}{x} \, dx \\ &= \frac{\pi}{16} \Big(3 \, \text{sgn}(1+a) + 3 \, \text{sgn}(1-a) - \text{sgn}(3+a) - \text{sgn}(3-a) \Big) \tag{1} \\ &= \begin{cases} 0 & \text{if} \ a >3 \\ - \frac{\pi}{16} & \text{if} \ a = 3 \\ - \frac{\pi}{8} & \text{if} \ 1<a <3 \\ \frac{\pi}{16} & \text{if} \ a = 1 \\ \frac{\pi}{4} & \text{if} \ a <1 \end{cases} \end{align}$$
Therefore,
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} f(e^{2ix})\, dx &= f(0) \left(\frac{\pi}{4} \right) + f'(0) \left(-\frac{\pi}{8} \right) + \sum_{n=2}^{\infty}\frac{f^{n}(0)}{n!} (0) \\ &= \frac{\pi}{8} \Big(2 f(0) - f'(0) \Big). \end{align}$$
$(1)$ $\int_{0}^{\infty} \frac{\sin (ax)}{x} \, dx = \frac{\pi}{2} \, \text{sgn}(a)$
As a simple example, let us consider the entire function $f(z) = e^{z}$.
Then $$\text{Re} \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{e^{2ix}} \, dx = \int_{0}^{\infty} \frac{\sin^{3} x}{x} \, e^{\cos (2x)} \cos (\sin 2x) \, dx = \frac{\pi}{8} \Big( 2(1)-1 \Big) = \frac{\pi}{8}.$$
Now let us consider the function $$ f(z) = \exp \left(a \, \frac{z-1}{z+1} \right)$$ for $a\ge 0$.
Since $\exp \left(a \, \frac{z-1}{z+1} \right)$ has an essential singularity at $z=-1$, we know that its Maclaurin series has a radius of convergence of $1$.
What we don't immediately know, and what is proving hard to ascertain, is if the series converges on the unit circle (excluding the point $z=-1$).
But assuming that it does we have
$$ \begin{align} \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp \left(a \, \frac{e^{2ix}-1}{e^{2ix}+1} \right) \, dx &= \text{Re} \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \, \exp (ia \tan x) \, dx \\ &= \int_{0}^{\infty} \frac{\sin^{3}(x)}{x} \cos(a \tan x) \, dx \\ &= \frac{\pi}{8} \Big(2 (e^{-a}) -2ae^{-a} \Big) \\ &= \frac{\pi}{4} (1-a) e^{-a}. \end{align}$$
