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Problem of the Week #253 - Jul 19, 2017

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Euge

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Jun 20, 2014
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Euge

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Jun 20, 2014
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Congratulations to Opalg for his correct solution. Honorable mention goes to fatimarose23 . Here is Opalg 's solution.


Substitute $s = t^{1/2}$, so that $t=s^2$ and $dt = 2s\,ds$. Then $$\int_0^\infty \frac{dt}{(1+t^2)t^{1/2}} = \int_0^\infty \frac{2s}{(1+s^4)s}ds = \int_0^\infty \frac{2}{(1+s^4)}ds.$$ To evaluate \(\displaystyle J = \int_0^\infty \frac{2}{(1+s^4)}ds\) by contour integration, use a contour consisting of the interval $[0,R]$, a quarter-circle from $R$ to $iR$ and then back down the imaginary axis to the origin (and then let $R\to\infty$). The integral of $\dfrac2{1+z^4}$ along the positive real axis is just $J$, the integral round the quarter-circle goes to $0$, and the integral down the imaginary axis is $-iJ$. So the integral round the whole contour is $(1-i)J.$

The only singularity inside the contour is at the fourth root of $-1$ at $z_0 = \dfrac{1+i}{\sqrt2}$. The residue there is $\dfrac2{4z^3}$ evaluated at $z_0$, namely $\dfrac{\sqrt2}{2(-1+i)}.$ So Cauchy's theorem tells us that $(1-i)J = \dfrac{2\pi i\sqrt2}{2(-1+i)}$. Therefore $$J = \frac{\pi i\sqrt2}{- (1-i)^2} = \frac{\pi i\sqrt2}{2i} = \frac{\pi}{\sqrt2}.$$
 
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