Calculate $\int_0^\infty \frac{\sin\left(x^2 - \arctan\left(\frac{1}{x^2}\right)\right)}{\sqrt{1 + x^4}} \, dx $

Question

Calculate $$\int_0^\infty \frac{\sin\left(x^2 - \arctan\left(\frac{1}{x^2}\right)\right)}{\sqrt{1 + x^4}} \, dx =? $$

$$\sin(\arctan(x^2)) = \frac{x^2}{\sqrt{1 + x^4}}; \quad \cos(\arctan(x^2)) = \frac{1}{\sqrt{1 + x^4}}; \quad \arctan(x^2) + \arctan\left(\frac{1}{x^2}\right) = \frac{\pi}{2}$$

$$\frac{{\sin(x^2 - \arctan(1/(x^2)))}}{{\sqrt{1 + x^4}}} = \frac{{x^2 \cdot \sin(x^2) - \cos(x^2)}}{{1 + x^4}}$$

We know that $$\int_0^1 e^{-ay} \cdot \cos(by) \, dy = \frac{a + b e^{-a} \cdot \sin(b) - a e^{-a} \cdot \cos(b)}{a^2 + b^2}$$

Now back to our original topic

$$\int_0^1 e^{-y} \cdot \cos(x^2 \cdot y) \, dy = \frac{1}{{1 + x^4}} + \frac{{x^2 \cdot \sin(x^2) - \cos(x^2)}}{{e(1 + x^4)}}$$

$$\iint\limits_{0}^{\infty} \int_0^1 e^{-y} \cdot \cos(x^2 y) \, dy \, dx = \int_0^{\infty} \frac{1}{1+x^4} \, dx + \frac{1}{e} \int_0^{\infty} \frac{{x^2 \cdot \sin(x^2) - \cos(x^2)}}{{1 + x^4}} \, dx$$

Answer 1

Continuing where OP left off. \begin{align*} I & = e \left( \int_{0}^{\infty} \int_{0}^{1} e^{-y}\cos(x^2y) \, \mathrm{d}y \, \mathrm{d}x -\int_{0}^{\infty} \frac{\mathrm{d}x}{1+x^4} \right)\\ & \overset{\color{purple}{u = x\sqrt{y}}}{=} e \left( \int_{0}^{1}\frac{e^{-y}}{\sqrt{y}} \int_{0}^{\infty} \cos(u^2) \, \mathrm{d}u \, \mathrm{d}y -\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right) \right)\\ & \overset{\color{purple}{v = \sqrt{y}}}{=} e \left(2\frac{\sqrt{\pi}}{2\sqrt{2}} \int_{0}^{1} e^{-v^2} \, \mathrm{d}v -\frac{\pi}{2\sqrt{2}} \right)\\ & = \boxed{ -\frac{e \pi \, \mathrm{erfc}(1)}{2 \sqrt{2}}} \end{align*} using that $\int_{0}^{\infty} \frac{t^{a-1}}{1+t^b}\mathrm{d}t = \frac{\pi}{b} \csc\left(\frac{\pi a}{b}\right)$ for $0<a<b$, the evaluation of the Fresnel integral $\int_{0}^{\infty}\cos(t^2)\, \mathrm{d}t = \frac{\sqrt{\pi}}{2\sqrt{2}}$, and the definition of the complementary error function $\mathrm{erfc}(z) = 1-\frac{2}{\sqrt{\pi}}\int_{0}^{z}e^{-t^2}\, \mathrm{d}t $.

Answer 2

An alternative solution. We define $$ I(a) = \int_{0}^{\infty}\frac{\cos(ax^2)}{1+x^4}\, \mathrm{d}x $$ noticing that the desired integral is $[-I'(a) - I(a)]\big\vert_{a=1}$. By Feynman's trick we get $$ I(a) - I''(a)=\int_{0}^{\infty} \cos(ax^2)\, \mathrm{d}x \overset{\color{purple}{x\sqrt{a}\to x}}{=} \frac{1}{\sqrt{a}}\int_{0}^{\infty}\cos(x^2) \, \mathrm{d}x = \frac{1}{2}\sqrt{\frac{\pi}{2a}} $$ using the evaluation of the Fresnel integral $\int_{0}^{\infty}\cos(t^2)\, \mathrm{d}t = \frac{\sqrt{\pi}}{2\sqrt{2}}$. The previous differential equation can be solved by variation of parameters, resulting in $$ I(a) = \int_{0}^{\infty}\frac{\cos(ax^2)}{1+x^4}\, \mathrm{d}x = \frac{\pi e^{-a} \left(e^{2 a} \text{erfc}\left(\sqrt{a}\right)+\text{erfi}\left(\sqrt{a}\right)+1\right)}{4 \sqrt{2}} $$ with erfc and erfi the complementary/imaginary error functions. Differentiating we get $$ I'(a)= -\int_{0}^{\infty}\frac{x^2\sin(ax^2)}{1+x^4}\, \mathrm{d}x = \frac{\pi e^{-a} \left(e^{2 a} \text{erfc}\left(\sqrt{a}\right)-\text{erfi}\left(\sqrt{a}\right)-1\right)}{4 \sqrt{2}} $$ so adding the equations for $I$ and $I'$ gives $$ \int_{0}^{\infty} \frac{x^2\sin(ax^2)-\cos(ax^2)}{1+x^4}\, \mathrm{d}x = -\frac{\pi e^a \text{erfc}\left(\sqrt{a}\right)}{2 \sqrt{2}} $$ and evaluating at $a=1$ gives the desired result.