Integrate a function using variable changement

Question

I did the integraton using Wolfram alpha but I wonder if is it feasible using either integration by part or by substitution

Answer 1

One out of many, many lovely proofs out there of this with double integration:

$$\left(\int_{-\infty}^\infty e^{-x^2}dx\right)^2=\int_{-\infty}^\infty e^{-x^2}dx\int_{-\infty}^\infty e^{-y^2}dy=\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2-y^2}dxdy\stackrel{\text{polar coor.}}=$$

$$=\int_0^{2\pi}d\theta\int_0^\infty \overbrace{r}^{\text{Jacobian!}}e^{-r^2}dr=\left.-\pi e^{-r^2}\right|_0^\infty=\pi$$

Now just take square roots

Answer 2

No. There is no analytic expression to $$ \int_{-\infty}^x e^{-x^2} dx \ .$$ However, using tricks one can prove $$ \int_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi} \ .$$

Denote $ I = \int_{-\infty}^{\infty} e^{-x^2} dx $. Then by changing names we have also $I = \int_{-\infty}^{\infty} e^{-y^2} dy$. Now $$ I^2 = \int_{-\infty}^{\infty} e^{-x^2} dx \int_{-\infty}^{\infty} e^{-y^2} dy = \iint e^{-(x^2 + y^2)}dxdy $$ integrating over the all plane $\mathbb{R}^2$. We now transform into polar coordinates: $$ x^2 + y^2 = r^2 \ , \quad x = r \cos \theta \mbox{ and } y = r \sin \theta $$ and using the Jacobian formula $dx dy = r dr d\theta$. Plugging it back to $I^2$ we get $$ I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} r dr d\theta = 2\pi \int_0^{\infty} e^{-r^2}r dr $$ It is easily shown that $$ \int e^{-r^2} r dr = \frac{-e^{-r^2}}{2}$$ and using Newton-Leibnitz formula $$ I^2 = 2\pi \left[ \frac{-e^{-r^2}}{2} \right]_0^{\infty} = 2\pi \left[ \frac{e^{-r^2}}{2} \right]^0_{\infty} = \pi \cdot (1 - 0) = \pi \ . $$ We got $I^2 = \pi$ and hence $I = +\sqrt{\pi}$ (since $I$ is clearly positive because $e^{-r^2} \ge 0$).