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How to compute $$\int\limits_0^\infty \frac{e^{-x}}{\sqrt{x}}\ dx?$$

I tried integration by parts, but it gave me division by 0. Should I substitute variable?

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    $\begingroup$ What did you choose for your integration by parts parameters? $\endgroup$ Apr 24 '19 at 18:42
  • $\begingroup$ Render $u=x^{1/2}$. Have you seen how to then integrate $e^{-u^2} du$? $\endgroup$ Apr 24 '19 at 18:44
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First substitute $x=y^2$ so your integral is $2\int_0^\infty e^{-y^2}dy$. Then pick your favourite proof this is $\sqrt{\pi}$. The first is by far the most common; the sixth doesn't even require the above substitution.

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Using the $u$-substitution $u=\sqrt{x}$, we find $$\int_0^\infty\frac{e^{-x}}{\sqrt{x}}\,dx=2\int_0^\infty e^{-u^2}\,du.$$ This latter integral is well known so that we have $$\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx=\sqrt{\pi}.$$

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What works here is to get this into the form of a Gaussian integral by letting $x = u^2$. Then $dx = 2u \,du$ and the integral becomes $$ \int_{u=0}^\infty \frac1 u e^{-u^2} 2u\,du = \int_{u=0}^\infty e^{-u^2} du = \sqrt{\pi} $$

The way that Gaussian integral is found is by first saying that it is half the integral from $-\infty$ to $\infty$, then saying that integral is the square root of the double integral $dx\,dy$ of $e^{-x^2} e^{-y^2}$, then changing to polar coordinates to get $$\int_{r=0}^\infty e^{-r^2} r\,dr\,d\theta$$ and because the $r$ from the transformation of $dy\,dx$ has come into play, this integral is easy.

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For $\Re[z] > 0, $ $$\Gamma(z) = \int_{0}^{\infty}t^{z-1}e^{-t}dt$$ For $z \in \mathbb N,$ $$\Gamma(z) = (z - 1)!$$

For $z \notin \mathbb Z,$ $$\Gamma(1-z)\Gamma(z) = \frac{\pi}{sin(\pi z)}$$

So, to solve $\int_{0}^{\infty}x^{-1/2}e^{-x}dx$, let $z = \frac{1}{2}$ in the equation above, thus we have $$\int_{0}^{\infty}x^{-1/2}e^{-x}dx = \Gamma(\frac{1}{2})$$

Now, since $z = \frac{1}{2}$, $$\Gamma(1 - \frac{1}{2})\Gamma(\frac{1}{2}) = \frac{\pi}{sin(\pi z)}$$ $$\therefore \space \Gamma(\frac{1}{2})^2 = \frac{\pi}{sin(\frac{\pi}{2})} = \pi$$ $$ \therefore \space \Gamma(\frac{1}{2}) = \sqrt{\pi}$$

Thus, we have that $$\int_{0}^{\infty}x^{-1/2}e^{-x}dx = \sqrt{\pi}. \quad \blacksquare$$

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  • $\begingroup$ Euler's Reflection Formula for the Gamma Function is quite tendious to prove (at least using elementary means)! It is kind of overdone to use this formula here ^^ $\endgroup$
    – mrtaurho
    Apr 24 '19 at 19:14
  • $\begingroup$ You can also use Legendre's duplication formula with the Beta function, which is much easier to prove $\endgroup$
    – Victoria M
    Apr 24 '19 at 19:20
  • $\begingroup$ @mrtaurho: Agreed... a cannon to kill a flea. $\endgroup$ Apr 24 '19 at 19:25
  • $\begingroup$ @VictoriaM Indeed. But this is still ridiculous heavy machinery for this easy problem ^^ $\endgroup$
    – mrtaurho
    Apr 24 '19 at 19:30

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