Interesting improper integral with 3rd order derivative.

This is an application of Leibniz's rule - ie, differentiating under the integral sign (a favourite trick of Richard Feynman, no less!)

As it stands, the integral looks (and is) pretty horrific. The trick is to differentiate wrt $n$ first ; the integrand is a cubic polynomial in $n$ , so its third derivative is just $6$ times the coefficient of $n^3$ in the integrand.

The integral we actually need is $I=\int_0^{\infty} \frac{6}{1+\pi^2 x^2} dx$ which can easily be worked out (the indefinite integral is just an inverse tangent) to be $\boxed3$ .

Please, see also Chris Lewis's solution. This is just a an expansion of it.

$\int_0^{\infty } \frac{\left(n+\frac{\sin (x)}{x}\right)^3}{\pi ^2 x^2+1} \, dx \Rightarrow \\ \frac{n^3}{2}-\frac{3}{2} e^{-1/\pi } \pi n^2+\frac{3 \pi n^2}{2}+\frac{3}{4} e^{-2/\pi } \pi ^2 n-\frac{3 \pi ^2 n}{4}+\frac{3 \pi n}{2}-\frac{1}{8} e^{-3/\pi } \pi ^3+\frac{3}{8} e^{-1/\pi } \pi ^3-\frac{\pi ^3}{4}+\frac{3 \pi }{8}$

$\frac{\partial \left(\frac{n^3}{2}-\frac{3}{2} e^{-1/\pi } \pi n^2+\frac{3 \pi n^2}{2}+\frac{3}{4} e^{-2/\pi } \pi ^2 n-\frac{3 \pi ^2 n}{4}+\frac{3 \pi n}{2}-\frac{1}{8} e^{-3/\pi } \pi ^3+\frac{3}{8} e^{-1/\pi } \pi ^3-\frac{\pi ^3}{4}+\frac{3 \pi }{8}\right)}{\partial n} \Rightarrow \\ \frac{3 n^2}{2}-3 e^{-1/\pi } \pi n+3 \pi n+\frac{3}{4} e^{-2/\pi } \pi ^2-\frac{3 \pi ^2}{4}+\frac{3 \pi }{2}$

$\frac{\partial \left(\frac{3 n^2}{2}-3 e^{-1/\pi } \pi n+3 \pi n+\frac{3}{4} e^{-2/\pi } \pi ^2-\frac{3 \pi ^2}{4}+\frac{3 \pi }{2}\right)}{\partial n} \Rightarrow 3 n-3 e^{-1/\pi } \pi +3 \pi$

$\frac{\partial \left(3 n-3 e^{-1/\pi } \pi +3 \pi \right)}{\partial n} \Rightarrow 3$

Starting with the expansion of the integrand, doing the problem as Chirs Lewis actually did the problem.

$\frac{n^3}{\pi ^2 x^2+1}+\frac{3 n^2 \sin (x)}{x \left(\pi ^2 x^2+1\right)}+\frac{3 n \sin ^2(x)}{x^2 \left(\pi ^2 x^2+1\right)}+\frac{\sin ^3(x)}{x^3 \left(\pi ^2 x^2+1\right)}$

$\frac{3 n^2}{\pi ^2 x^2+1}+\frac{6 n \sin (x)}{x \left(\pi ^2 x^2+1\right)}+\frac{3 \sin ^2(x)}{x^2 \left(\pi ^2 x^2+1\right)}$

$\frac{6}{\pi ^2 x^2+1}$

$6\,\int \frac{1}{\pi ^2 x^2+1} \, dx \Rightarrow 6\frac{\tan ^{-1}(\pi x)}{\pi }|_{x=0}^{x=\infty} \Rightarrow 6(\frac12-0) \Rightarrow 3$

${\color{#EC7300} \text{Added 2019 June 6 1330Z (D-Day 75th anniversary).}}$

The indefinite integral:

$\frac{1}{4 \pi}(4 n \sinh \left(\frac{1}{\pi }\right) \text{Ci}\left(\frac{i}{\pi }-x\right)+4 n \sinh \left(\frac{1}{\pi }\right) \text{Ci}\left(\frac{i}{\pi }+x\right)-i \cosh \left(\frac{2}{\pi }\right) \text{Ci}\left(\frac{2 i}{\pi }+2 x\right)+i \cosh \left(\frac{2}{\pi }\right) \text{Ci}\left(\frac{2 i}{\pi }-2 x\right)+ \\ 4 n^2 \tan ^{-1}(\pi x)+4 i n \cosh \left(\frac{1}{\pi }\right) \text{Si}\left(\frac{i}{\pi }-x\right)+4 i n \cosh \left(\frac{1}{\pi }\right) \text{Si}\left(\frac{i}{\pi }+x\right)+\sinh \left(\frac{2}{\pi }\right) \text{Si}\left(\frac{2 i}{\pi }+2 x\right)-\sinh \left(\frac{2}{\pi }\right) \text{Si}\left(\frac{2 i}{\pi }-2 x\right)+ \\ 2 \tan ^{-1}(\pi x))$

Assuming that $z\in \mathbb{R}_{>\, 0}\lor \Im(z)\neq 0,\, \text{Ci}(z)=-\int_z^{\infty } \frac{\cos (t)}{t} \, dt$ .

$\text{Si}(z) = \frac{1}{2} (\pi -2 \int_z^{\infty } \frac{\sin (t)}{t} \, dt)$ .

These special functions can be developed by taking the Maclaurin series for the respective trigonometric series, doing the division by $t$ and doing the integration from z to infinity on each term individually and then reassembling the new series. This is done for otherwise undoable integrations.

Evaluating the indefinite integral at 0 gives:

$\frac{-8 n \text{Shi}\left(\frac{1}{\pi }\right) \cosh \left(\frac{1}{\pi }\right)+8 \text{Ci}\left(\frac{i}{\pi }\right) n \sinh \left(\frac{1}{\pi }\right)}{4 \pi }$

Evaluating the indefinite integral at $\infty$ gives:

$\frac{2 \pi n^2+4 i \pi n \sinh \left(\frac{1}{\pi }\right)+\pi +\pi \sinh \left(\frac{2}{\pi }\right)-\pi \cosh \left(\frac{2}{\pi }\right)}{4 \pi }$

${\large\color{#D61F06} \text{The point is: this is unnecessary! See Chris Lewis's much simpler solution.}}$