How is integral calculated?
$$\int_ 0^\infty e^{-x^2}~dx $$
Asked
Active
Viewed 141 times
1
-
1You can't do it in terms of a finite number of elementary functions, no matter what the method. – DonAntonio Apr 09 '13 at 18:04
-
2If the integral is from $0$ to $\infty$, then the solution is easy though. – Lord Soth Apr 09 '13 at 18:05
-
1This has been asked many times before, e.g. this. – J. M. ain't a mathematician Apr 09 '13 at 18:09
-
Question edited (added 0 to $\infty$) – TheNotMe Apr 09 '13 at 18:13
-
Make the change of variables $t^2=u$ and then use gamma function. – Mhenni Benghorbal Apr 09 '13 at 18:34
4 Answers
1
The standard way to do this integral is to evaluate its square. That is, let
$$I = \int_0^{\infty} dx \: e^{-x^2}$$
Then
$$I^2 = \int_0^{\infty} dx \: e^{-x^2} \int_0^{\infty} dy \: e^{-y^2} = \int_0^{\infty} dx \:\int_0^{\infty} dy \: e^{-(x^2+y^2)}$$
Now the trick is to convert to polar coordinates: $dx dy = r dr d\theta$, $r \in [0,\infty)$, $\theta \in [0, 2 \pi)$:
$$I^2 = \int_0^{\infty} dr \,r\, e^{-r^2} \int_0^{2 \pi} d\theta = 2 \pi \frac{1}{2} = \pi$$
Therefore
$$I = \sqrt{\pi}$$
Ron Gordon
- 138,521
0
Neither. The function has no antiderivative that can be expressed in terms of elementary functions, so both those methods fall down.
If you wish to evaluate the definite integral over a particular interval, though, you've got a shot.
Cameron Buie
- 102,994
0
See Gaussian integral and Error function. It is $\sqrt \pi $ on $(- \infty, +\infty) $ . Indefinite could not be expressed in elementary functions.
igumnov
- 809