Integration problem solving without contour integration 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 $\alpha=\frac{(4n-1)\pi\sqrt{2}}{4\beta}$
 A: Switching to integration parameter $t=\frac{\alpha}{\beta}x$ gives $$F(\alpha,\beta)=\displaystyle\int_0^\infty\frac{\cos(\beta t)}{(\frac{\beta}{\alpha}t)^4+\beta^4}\,\frac{\beta}{\alpha}\,dt=\frac{\alpha^3}{\beta^3}F(\beta,\alpha)$$
A: 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. 
