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I want to show that $\int_{0}^{\infty}\frac{\sqrt{x}}{1+x^2}\,dx = \frac \pi {\sqrt 2}$.

I'm following much the same approach as in this answered question (i.e., integrating around the keyhole contour in the complex plane), $\int_{0}^{\infty}\frac{\sqrt{x}}{1+x^2}\,dx$

but have gotten stuck at the same place as the original poster. Namely, I want to calculate $2\pi i \sum \text{Residues}$ of $f(z) = \frac{\sqrt z}{z^2 + 1}$. Since there are simple poles at $z = i, -i$ inside the contour. my calculations seem to keep getting $\frac 1 {\sqrt 2}$ as the sum of the residues, and I suspect I'm not using the branch cut correctly. However, I'm not sure how to do the residue calculations w.r.t the branch cut correctly.

greg115
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  • https://math.stackexchange.com/questions/97653/what-contour-should-be-used-to-evaluate-int-0-infty-frac-sqrtt1t2-dt – Sine of the Time Nov 22 '22 at 22:30
  • @SineoftheTime I read that question but was not sure which part could apply here, since the OP there use a different contour (not the keyhole contour) – greg115 Nov 22 '22 at 22:52

1 Answers1

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Substitute $t=\sqrt x$ in the integral. Then write the integrand as $$\frac{2t^2}{1+t^4}=\frac{2t^2}{(t^2-\sqrt 2{\space} t+1)(t^2+\sqrt 2 {\space}t+1)}$$ Break the integrand apart by partial fractions. Integrate from $t=0$ to $t=b.$ Take the limit as $b \rightarrow \infty$

P. Lawrence
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