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Problem of the Week #257 - Sep 26, 2017

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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
Here is this week's POTW:

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Using contour integration, prove

$$\int_0^\infty \sin(x^\alpha)\, dx = \sin\left(\frac{\pi}{2\alpha}\right)\,\Gamma\!\left(1 + \frac{1}{\alpha}\right),\quad \alpha > 1.$$


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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
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This week's problem was solved by Opalg . You can read his solution below.


Integrate the function $f(z) = e^{iz^\alpha}$ over a "pizza slice" contour consisting of (1) the x-axis from $0$ to $r$, (2) a circular arc of radius $r$ from $r$ to $re^{i\pi/(2\alpha)}$, (3) a straight line from $re^{i\pi/(2\alpha)}$ to $0$, and then let $r\to\infty$. Call the three integrals $I_1$, $I_2$, $I_3$.

Then \(\displaystyle I_1 = \int_0^\infty e^{ix^\alpha}dx = \int_0^\infty (\cos(x^\alpha) + i\sin(x^\alpha))\,dx.\)

For $I_2$, use the parametrisation $z = re^{i\theta}$. Then $$|f(z)| = \bigl|\exp\bigl(i(re^{i\theta})^\alpha\bigr)\bigr| = \bigl|\exp\bigl(ir^\alpha e^{i\alpha\theta}\bigr)\bigr| = \bigl|\exp\bigl(ir^\alpha (\cos(\alpha\theta) + i\sin(\alpha\theta))\bigr)\bigr| = \bigl|\exp\bigl(ir^\alpha \cos(\alpha\theta) - r^\alpha\sin(\alpha\theta))\bigr)\bigr| \leqslant \bigl|e^{-r^\alpha\sin(\alpha\theta)}\bigr|.$$ But for $0\leqslant \theta \leqslant \pi/(2\alpha)$, $\sin(\alpha\theta) \geqslant k\alpha\theta$ for some positive constant $k$ (in fact, you could take $k = 2/\pi$). Therefore, on the contour of $I_2$, $|f(z)| \leqslant \exp(-kr^\alpha\theta)$, and so $$|I_2| \leqslant \int_0^{\pi/(2\alpha)}\exp(-kr^\alpha\theta)r\,d\theta = \Bigl[\frac{-r}{kr^\alpha}\exp(-kr^\alpha\theta)\Bigr]_0^{\pi/(2\alpha)} \leqslant \frac1{kr^{\alpha-1}} \to0 \text{ as } r\to\infty.$$

So $I_2 = 0$. That was a delicate and somewhat messy calculation, but I don't see how to shorten it.

For $I_3$ let $z = se^{i\pi/(2\alpha)}$. Then $$I_3 = \int_\infty^0 e^{-s^\alpha}e^{i\pi/(2\alpha)}ds = -e^{i\pi/(2\alpha)}\int_0^\infty e^{-s^\alpha}ds.$$ Substitute $t = s^\alpha$ (so that $dt = \alpha s^{\alpha-1}dt = \alpha t^{1 - \frac1\alpha}\,ds$) and integrate by parts to get $$\int_0^\infty e^{-s^\alpha}ds = \int_0^\infty e^{-t} \tfrac1\alpha t^{ \frac1\alpha - 1}\,dt = \int_0^\infty t^{1/\alpha}e^{-t}dt = \Gamma\bigl(1 + \tfrac1\alpha \bigr).$$ Hence $I_3 = -e^{i\pi/(2\alpha)}\Gamma\bigl(1 + \frac1\alpha \bigr)$.

The function $f(z)$ has no zeros inside the contour (or anywhere else for that matter), and so $I_1 + I_2 + I_3 = 0$. Therefore $$ \int_0^\infty (\cos(x^\alpha) + i\sin(x^\alpha))\,dx = e^{i\pi/(2\alpha)}\Gamma\bigl(1 + \tfrac1\alpha \bigr).$$ Finally, take the imaginary part to get \(\displaystyle \int_0^\infty \sin(x^\alpha)\,dx = \sin\bigl(\tfrac\pi{2\alpha}\bigr)\Gamma\bigl(1 + \tfrac1\alpha \bigr).\)
 
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