I'm trying to solve this problem:
By differentiating the definition of the Fourier sine transform $\tilde f_s(w)$ of the function $f(t) = t^{−1/2}$ with respect to ω, and then integrating the resulting expression by parts, find an elementary differential equation satisfied by $\tilde f_s(w)$. Hence show that this function is its own Fourier sine transform, i.e. $\tilde f_s(w)$ = Af(w), where A is a constant. Assume that the limit as x → ∞ of $x^{1/2} sin (\alpha x)$ can be taken as zero.
My attempt:
Fourier sine transform of f is given by
$$\sqrt{\frac{2}{\pi}} \int^{\infty}_0 t^{-1/2}sin(wt)\, dt $$
Differentiating this w.r.t w gives:
$$\sqrt{\frac{2}{\pi}} \int^{\infty}_0 t^{1/2}cos(wt)\, dt $$
integrating by parts:
$$\sqrt{\frac{2}{\pi}} [\frac{t^{1/2}sin(wt)}{w} |^{\infty}_0 – \int^{\infty}_0 \frac{sin(wt)}{2wt^{1/2}}\, dt] $$
by assumption as x → ∞ of $x^{1/2} sin (\alpha x)$ can be taken as zero:
$$-\sqrt{\frac{2}{\pi}} \int^{\infty}_0 \frac{sin(wt)}{2wt^{1/2}}\, dt $$
My question is how do I take this integral?
Thanks!
Best Answer
Let $$F(w) = \int_0^\infty t^{-1/2} \sin (w t)\, dt$$
$$F(w) = w^{-1/2} \int_0^{\infty} w^{-1/2}t^{-1/2} \sin wt \, w dt$$
Let $q=wt$.
$$F(w) = w^{-1/2} \int_0^\infty q^{-1/2} \sin q \, dq = \text{const.}\cdot w^{-1/2}.$$
It is known that $\int_0^\infty q^{-1/2} \sin q \, dq = \sqrt{\frac{\pi}{2}}$, but we don't need to evaluate this integral to show that $t^{-1/2}$ is its own Fourier transform (up-to-a-constant).
UPDATE:
One way to compute $\int_0^\infty q^{-1/2} \sin q\, dq = 2\int_0^\infty \sin p^2 \, dp, $ is to consider the contour integral $$ \oint_C e^{i z^2} \, dz$$ with C the segment $0$ to $R$ along the $x$-axis, a quarter circle of radius $R$ from $R$ to $z^\star=Re^{i\pi/4}$ and a segment from $z^\star$ to the origin (a slice (1/8) of a pie of radius $R$).
Then $$0=\oint_C e^{i z^2} \, dz = \int_0^\infty e^{ip^2} \, dp - e^{i\pi/4} \int_0^\infty e^{-y^2} \, dy + \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta,$$ and
$$ \begin{aligned} \left| \int_{\Gamma_R} \exp (Re^{i2\theta})R i e^{i\theta}\,d\theta \right| &\le \int_{\Gamma_R} \left| \exp (R^2e^{i2\theta})R i e^{i\theta}\right| \,d\theta \\ &= \int_0^{\pi/4} Re^{-R^2\sin 2\theta} \,d\theta \\&<\int_0^{\pi/4} Re^{- \frac{4}{\pi} R^2\theta} \, d\theta \\&=\left(1-e^{-R^2} \right)\frac{\pi}{4R} \to 0.\end{aligned} $$
So
$$ \int_0^\infty \cos p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
$$ \int_0^\infty \sin p^2 \, dp = \frac{1}{2} \sqrt{\frac{\pi}{2}},$$
and the Fourier sine transform of $t^{-1/2}$ is $w^{-1/2}$.
$$t^{-1/2} \iff w^{-1/2}.$$