Integration problem solving without contour integration

Question

Can the following question be solved without using contour integration.

$F:(0,\infty)\times (0,\infty)\to \Bbb R$ be given by

$F(\alpha,\beta)=\displaystyle\int_0^\infty\frac{\cos(\alpha x)}{x^4+\beta^4}\,dx$

Show that

$\frac{F(\alpha,\beta)}{F(\beta,\alpha)}=\frac{\alpha^3}{\beta^3}$ as long as there is no positive integer n such that $\alpha=\frac{(4n-1)\pi\sqrt{2}}{4\beta}$

Answer 1

To answer your question concerning why the quotient $$\frac{F(\alpha, \beta)}{F(\beta, \alpha)} = \frac{\alpha^3}{\beta^3},$$ is only defined provided $$\alpha \neq \frac{(4n - 1) \pi \sqrt{2}}{4b},$$ where $n \in \mathbb{N}$, it is helpful if one has the explicit expression for $F$ at hand.

Using the Residue Theorem a value for $F(\alpha, \beta)$ can be found. It is given by (for a proof of this, see here) $$F (\alpha, \beta) = \frac{\pi}{2 \alpha^3} \exp \left (-\frac{\alpha \beta}{\sqrt{2}} \right ) \sin \left (\frac{\alpha \beta}{\sqrt{2}} + \frac{\pi}{4} \right ), \qquad \alpha, \beta > 0.$$ Similarly, $$F (\beta, \alpha) = \frac{\pi}{2 \beta^3} \exp \left (-\frac{\alpha \beta}{\sqrt{2}} \right ) \sin \left (\frac{\alpha \beta}{\sqrt{2}} + \frac{\pi}{4} \right ), \qquad \alpha, \beta > 0.$$

If the required quotient is to be defined, clearly $F (\beta, \alpha) \neq 0$. Thus the sine term appearing in the expression for $F(\beta, \alpha)$ must be non-zero. Thus $$\frac{\alpha \beta}{\sqrt{2}} + \frac{\pi}{4} \neq n \pi \quad \text{or} \quad \alpha \neq \frac{(4n - 1) \pi \sqrt{2}}{4 \beta},$$ where $n \in \mathbb{N}$, as required.

Answer 2

Switching to integration parameter $t=\frac{\alpha}{\beta}x$ gives $$F(\alpha,\beta)=\displaystyle\int_0^\infty\frac{\cos(\beta t)}{(\frac{\beta}{\alpha}t)^4+\beta^4}\,\frac{\beta}{\alpha}\,dt=\frac{\alpha^3}{\beta^3}F(\beta,\alpha)$$