How to evaluate $\int_{-\infty}^{\infty} \frac{\cos((b) (r) (u) (h^2))}{(h^2) (r) (u) (b)} (h) \tan^{-1}(h) \, dh$

Question

How to evaluate \begin{align} \int_{-\infty}^{\infty} \frac{\cos((b) (r) (u) (h^2))}{(h^2) (r) (u) (b)} (h) \tan^{-1}(h) \, dh \end{align}

At the first sight the question looks like joke or troll but I'm asking it seriously

$$\int_{-\infty}^{\infty} \frac{\cos(b r u h^2)}{h^2 r u b} \cdot h \cdot \arctan(h) \, dh$$

Simplified:

$$2 \int_{0}^{\infty} \frac{\cos(b r u h^2)}{h^2 r u b} \cdot h \cdot \arctan(h) \, dh$$

With a substitution

$$\frac{1}{2rub} \int_{0}^{\infty} \frac{\cos(b r u x)}{x} \cdot \arctan(\sqrt{x}) \, dx$$

$$\frac{1}{2rub} \int_{0}^{\infty} \frac{\cos(b r u x)}{x} \cdot \arctan(\sqrt{x}) \, dx$$

$$u = \arctan(\sqrt{x} \quad \text{and} \quad dv = \frac{\cos(b r u x)}{x} \, dx$$

$$du = \frac{1}{\sqrt{x} (1 + x)} \, dx \quad \text{and} \quad v = \int \frac{\cos(b r u x)}{x} \, dx$$

Now with IBM

$\int u \, dv = uv - \int v \, du$

$$\frac{1}{2rub} \left[\left(\arctan(\sqrt{x}) \cdot \int \frac{\cos(b r u x)}{x} \, dx\right) - \int \left(\int \frac{\cos(b r u x)}{x} \, dx\right) \cdot \frac{1}{\sqrt{x} (1 + x)} \, dx\right]$$

$$\frac{1}{2rub} \left[\arctan(\sqrt{x}) \cdot v - \int v \, du\right]$$

Answer 1

As @Benjamin Wang commented, let $a=bru$ to make the integral $$I=\frac 2 a\int_0^\infty \frac 1 h \cos \left(a h^2\right)\,\tan ^{-1}(h)\,dh$$

Integration by parts $$u=\tan ^{-1}(h) \qquad \implies \qquad du=\frac{dh}{h^2+1}$$ $$dv=\frac 1 h \cos \left(a h^2\right)\,dh\qquad \implies \qquad v=\frac 12 \text{Ci}\left(a h^2\right)$$

$$\int_0^\infty u\,v\,dh=0$$ So $$I=-\frac 1 a\int_0^\infty \frac{\text{Ci}\left(a h^2\right)}{h^2+1}\,dh$$ I do not know what could be the antiderivative but the definite integral involves hypergeometric functions.

Mathematica returns $$a\,I=-\frac{\pi }{2}\, \text{Ci}(a)+\sqrt{2 \pi a} \,\,\, _2F_3\left(\frac{1}{4},1;\frac{3}{4},\frac{5}{4},\frac{5}{4};-\frac{a^2}{4}\right)-$$ $$\frac{2}{9}a \sqrt{2 \pi a} \,\,\, _2F_3\left(\frac{3}{4},1;\frac{5}{4},\frac{7}{4},\frac{7}{4};-\frac{a^2}{4}\right)$$

I suppose that you did not expect a so complicated result.