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I'm thinking of using the residue theorem. The residues are at $\pm i$ and I know how to calculate them but I'm stuck on finding a contour that fits? How do you find the correct contour in general?

mhighwood
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  • How about half of circle of diameter $0$ and $R$? – Kroki Jun 13 '23 at 13:46
  • The antiderivative exists in elementary functions: $\ln$ and $\arctan$. – zkutch Jun 13 '23 at 13:47
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    $f(t)=1/(1+t^2)$ has the Mellin transform, $$\phi(s)=\int_0^{\infty}t^{s-1}\frac{1}{1+t^2}\ dt=\frac{\pi}{2}\csc\left(\frac{\pi}{2}s\right)$$ take $s=3/2$, $$\phi(3/2)=\frac{\pi}{2}\csc\left(\frac{3\pi}{4}\right)=\frac{\pi}{\sqrt 2}.$$ – bob Jun 13 '23 at 14:31
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    The transformation can be proved by first evaluating the Mellin transform of $1/(1+t)$ using the substitution $1+t=1/(1-x)$ and some Beta function properties. Then use the relation, $$\mathcal{M}{f(t^a);s}=\frac{\phi(s/a)}{a}.$$ – bob Jun 13 '23 at 14:38

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You can compute the integral without the residus theorem. Indeed,

\begin{align} \int_0^\infty \frac{\sqrt t}{1 + t^2}\mathrm d t &= \int_0^1 \frac{\sqrt t}{1 + t^2}\mathrm d t + \int_1^\infty \frac{\sqrt t}{1 + t^2}\mathrm d t\\ &= \int_0^1 \frac{\sqrt t}{1 + t^2}\mathrm d t + \int_0^1 \frac{\sqrt {\frac1u}}{1 + \frac1{u^2}} \frac1{u^2}\mathrm d t\\ &= \int_{0}^1 \frac{\sqrt t + \frac1{\sqrt t}}{1+t^2}\mathrm dt\\ &= 2\int_{0}^1 \frac{1+t^2}{1+t^4}\mathrm d t\\ &= 2 \left[\frac1{\sqrt 2}\left(\arctan \left(t\sqrt 2 + 1\right) + \arctan \left(t\sqrt 2 - 1\right)\right)\right]\\ &= \ldots \end{align}

$$$$

Kroki
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