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Consider the following integral $$\int_0^\infty \frac{e^{-x}}{\sqrt{x}}dx=\sqrt{\pi}$$

This can be evaluated using contour integration methods. A similar question was asked before (unfortunately I could not find the link), however the person who answered did not provide the contour but said "Choose a suitable contour noting that $z=0$ is a branch point". Unfortunately I could not find a suitable contour for this problem with this integrand. The problem is probably that this function doesn't have any useful poles.

Could someone indicate what kind of contour works for this problem? You can even add poles if it helps to somehow evaluate above but without changing the integrand too much.

Gautam Shenoy
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  • I doubt there is a straight forward method because $\sqrt{\pi}$ can't be deduced from poles evaluation. The evaluation of the integral depends on $$\int^{\infty}_{-\infty} e^{-x^2} = \sqrt{\pi}$$ after substitution. – Zaid Alyafeai May 22 '17 at 19:19
  • Seems like there may be easier ways to evaluate that integral, but a suitable contour should be a "keyhole" type. Big outer circle of radius $R$, approach the origin above the positive $x$-axis, small circle of radius $\epsilon$ around the origin, back out to the outer circle below the positive $x$-axis. – sharding4 May 22 '17 at 19:21
  • That contour will fail to yield anything useful since $e^{-z}$ diverges as $\text{Re}(z)\to -\infty$. – Mark Viola May 22 '17 at 19:28
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    https://math.stackexchange.com/q/1266856/221811 has references. – Chappers May 22 '17 at 19:35
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    See the answer posted by @t.b. HERE for another way forward. – Mark Viola May 22 '17 at 19:46
  • @ Everyone: Thanks guys. What I was referring to was an answer someone had posted to this problem in some other post saying that "You can choose a suitable contour...". Guess the answer to this problem is, "Not this way". I recall reading in a book that the above integral was assumed unsolvable via contour methods until Polya found a way. – Gautam Shenoy May 23 '17 at 02:24

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