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Problem of the Week #196 - March 1, 2016

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Euge

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Jun 20, 2014
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Euge

MHB Global Moderator
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Jun 20, 2014
1,892
No one answered this week's problem. You can read my solution below.



Breaking up the integral over $[0,1/2]$ and $[1/2,1]$, and using the inequalities $t \le 1 - t$ for $t\in [0,1/2]$ and $t \ge 1 - t$ for $t\in [1/2,1]$, we find that the integral is dominated by

$$M_1\int_0^{1/2}t^{-3/4}\, dt + M_2 \int_{1/2}^1 (1 - t)^{-3/4}\, dt$$

where $M_1$ and $M_2$ are the maxima of $(16 - 7t)^{-5/4}$ over $[0,1/2]$ and $[1/2,1]$, respectively. Since $\int_0^{1/2} t^{-5/4}\, dt = \int_{1/2}^1 (1 - t)^{-5/4}\, dt < \infty$, it follows that our integral converges.

The calculation turns out to

$$\frac{\sqrt{\pi}}{6}\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$$

For let

$$u = \frac{9t}{16 - 7t}$$

Then $u = 0$ when $t = 0$ and $u = 1$ when $t = 1$. Furthermore,

$$t = \frac{16u}{9 + 7u},\qquad 1 - t = \frac{9(1 - u)}{9 + 7u},\qquad 16 - 7t = \frac{144}{9 + 7u},\qquad dt = \frac{144}{(9 + 7u)^2}\, du$$

and thus

$$\int_0^1 \frac{t^{-1/2}(1-t)^{-1/4}}{(16 - 7t)^{5/4}}\, dt = \int_0^1 \left[\frac{16u}{9 + 7u}\right]^{-1/2}\left[\frac{9(1-u)}{9+7u}\right]^{-1/4}\frac{(9+7u)^{5/4}}{144^{5/4}}\frac{144}{(9+7u)^2}\, du = 16^{-1/2}9^{-1/4}144^{-1/4}\int_0^1 u^{-1/2}(1-u)^{-1/4}$$
$$ = \frac{1}{24}B\left(\frac{1}{2},\frac{3}{4}\right)$$
$$=\frac{1}{24}\frac{\Gamma\left(\frac{1}{2}\right)\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{5}{4}\right)}$$
$$= \frac{1}{24}\frac{\sqrt{\pi}\Gamma\left(\frac{3}{4}\right)}{\frac{1}{4}\Gamma\left(\frac{1}{4}\right)}$$
$$= \frac{\sqrt{\pi}}{6}\frac{\Gamma\left(\frac{3}{4}\right)}{\Gamma\left(\frac{1}{4}\right)}$$
 
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