# $\int\limits_0^\infty \frac{e^{-x}}{\sqrt{x}} dx$

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?

• What did you choose for your integration by parts parameters? Apr 24 '19 at 18:42
• Render $u=x^{1/2}$. Have you seen how to then integrate $e^{-u^2} du$? Apr 24 '19 at 18:44

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.

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 $$\int_0^\infty \frac{e^{-x}}{\sqrt x}\,dx=\sqrt{\pi}.$$

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.

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

• 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 ^^ Apr 24 '19 at 19:14
• You can also use Legendre's duplication formula with the Beta function, which is much easier to prove Apr 24 '19 at 19:20
• @mrtaurho: Agreed... a cannon to kill a flea. Apr 24 '19 at 19:25
• @VictoriaM Indeed. But this is still ridiculous heavy machinery for this easy problem ^^ Apr 24 '19 at 19:30