MHB What is the solution to POTW #265 - Feb 13, 2018?

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Here is this week's POTW:

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Let $F : (0, \infty)\times (0,\infty) \to \Bbb R$ be given by
$$F(\alpha, \beta) = \int_0^\infty \frac{\cos(\alpha x)}{x^4 + \beta^4}\, dx$$ 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 $\alpha = \dfrac{(4n-1)\pi\sqrt{2}}{4\beta}$.-----

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Hi MHB community,

In the original posting of this week’s POTW I accidentally left out a necessary condition on $\alpha$ and $\beta$. I have made the correction.

 

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Congratulations to Opalg for his correct solution. You can read his solution below.

Spoiler

Let $$J = \oint_D\frac{e^{i\alpha z}}{z^4 + \beta^4}dz$$, where $D$ is the usual contour consisting of the interval $[-R,R]$ of the real axis, and the semicircle form $R$ to $-R$ taken anticlockwise in the upper half-plane. In the limit, as $R\to\infty$, the integral round the semicircle goes to $0$ by Jordan's lemma; and the integral along the axis becomes $$\int_{-\infty}^\infty \frac{\cos(\alpha x) + i\sin(\alpha x)}{x^4 + \beta^4}dx = 2\int_0^\infty \frac{\cos(\alpha x)}{x^4 + \beta^4}dx $$ (because $\cos(\alpha x)$ is an even function and $\sin(\alpha x)$ is an odd function). So $J = 2F(\alpha,\beta)$.

The function $$ \frac{e^{i\alpha z}}{z^4 + \beta^4}$$ has two simple poles inside $D$, at the points $\beta\omega$ and $\beta\omega^3$, where $\omega = e^{i\pi/4} = \dfrac{1+i}{\sqrt2}$.

The formula for the residue of a quotient of two analytic functions at a simple zero of the denominator says that the residue of $$ \frac{e^{i\alpha z}}{z^4 + \beta^4}$$ at such a point is $$ \frac{e^{i\alpha z}}{4z^3}$$. So the residues at $z = \beta\omega$ and $z = \beta\omega^3$ are $$\operatorname{Res}(\beta\omega) = \frac{e^{i\alpha\beta\omega}}{4\beta^3\omega^3} = \frac{-ie^{i\alpha\beta\omega}}{4\beta^3\omega}, \qquad \operatorname{Res}(\beta\omega^3) = \frac{e^{i\alpha\beta\omega^3}}{4\beta^3\omega^9} = \frac{e^{-\alpha\beta\omega}}{4\beta^3\omega}.$$ It follows from Cauchy's theorem that $$2F(\alpha,\beta) = \frac{2\pi i}{4\beta^3\omega}\bigl(-ie^{i\alpha\beta\omega} + e^{-\alpha\beta\omega}\bigr).$$

If $\alpha$ and $\beta$ are interchanged then the only effect on that last expression is that the $\beta^3$ in the denominator becomes $\alpha^3$. It follows that $\dfrac{F(\alpha,\beta)}{F(\beta,\alpha)} = \dfrac{\alpha^3}{\beta^3}.$

The only thing that can go wrong with that is that $F(\alpha,\beta)$ might be zero. In that case, $F(\beta,\alpha)$ will also be zero, and their quotient will not be defined. So we need to investigate the possibility that $ie^{i\alpha\beta\omega} = e^{-\alpha\beta\omega}$.

The exponential function has period $2\pi i$. So two complex numbers will have the same exponential if they differ by an integer multiple of $2\pi i$. But $ie^{i\alpha\beta\omega} = e^{i\pi/2}e^{i\alpha\beta\omega} = e^{i(\alpha\beta\omega + \pi/2)}$. Thus the condition for the result to fail is $$i(\alpha\beta\omega + \tfrac\pi2) = -\alpha\beta\omega + 2n\pi i = i^2\alpha\beta\omega + 2n\pi i,$$ $$\alpha\beta\omega + \tfrac\pi2 = i\alpha\beta\omega + 2n\pi,$$ $$(1-i)\alpha\beta\omega = \bigl(2n-\tfrac12\bigr)\pi.$$ But $\omega = \dfrac{1+i}{\sqrt2}$. So $(1-i)\omega = \dfrac{(1-i)(1+i)}{\sqrt2} = \sqrt2$, and the condition becomes $$\alpha\beta\sqrt2 = \frac{(4n-1)\pi}2,$$ which is equivalent to the given condition for the result to fail.