$\int_{- \infty}^{\infty} e^{- x^{2}} dx$

By Anonymous on Saturday, March 24, 2001 - 01:17 pm :

I have been looking at a way of evaluating

\int_{- \infty}^{\infty} e^{- x^{2}} dx

. However, I am getting stuck when it comes to using double integration in the polar plane. Can anyone help?

By Kerwin Hui (Kwkh2) on Saturday, March 24, 2001 - 01:54 pm :

Amars,

Let $I = \int_{- \infty}^{\infty} e^{- x^{2}} dx$

Then,

$I^{2} = \int_{- \infty}^{\infty} e^{- x^{2}} dx \int_{- \infty}^{\infty} e^{- y^{2}} dy$

$= \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} e^{- x^{2}} e^{- y^{2}} dy dx$

$= \int_{0}^{2 π} \int_{0}^{\infty} e^{- r^{2}} r dr d θ$

$= π$

so $I = \sqrt{π}$

Hence $\int_{- \infty}^{\infty} e^{- x^{2}} dx = - \sqrt{π}$

Kerwin

By Emma McCaughan (Emma) on Monday, March 26, 2001 - 09:50 am :

For another explanation, see Integral of Gaussian function .

By Amars Birdi (P2769) on Wednesday, March 28, 2001 - 05:05 pm : Thanks for your replies, but I still can't see why, when converting from cartesian to polar co-ordinates,

dx dy = r dr d θ

. This has something to do with areas - but what? -Thanks!

By Amars Birdi (P2769) on Wednesday, March 28, 2001 - 05:23 pm :

Also, why is it that when you multiply two separate integrals (one with respect to

y

and the other with respect to

x

) you are allowed to end up with an integral of an integral? More to the point, when you integrate

z = e^{- x^{2}}

this is an integral with respect to

x

, when you integrate (another variable)

= e^{- y^{2}}

this is an integral with respect to

y

. What does it MEAN to multiply these two integrals - are we multiplying the numerical values of each integral OR are we multiplying two separate integrals in separate variables (I guess this is connected with what a double integral is)

By David Loeffler (P865) on Wednesday, March 28, 2001 - 08:55 pm :

Amars,

As regards the last question, we need a fundamental result called Fubini's theorem, which states that the multiple integral

\int \int F (x, y) dx dy

is equal to the ''iterated integral''

$\int (\int F (x, y) dx) dy$

where we first evaluate the integral with respect to $x$ , treating $y$ as a parameter, then integrate this with respect to $y$ . (I've written this for two variables but it extends to any number in the obvious manner.)

Of course if $F (x, y) = f (x) g (y)$ then this becomes the product of the two integrals $\int f (x) dx$ and $\int g (y) dy$ , which is the result we need to evaluate the $e^{- x^{2}}$ integral.
David

By Michael Doré (Md285) on Wednesday, March 28, 2001 - 10:23 pm :

Here is an outline of another (non-standard) way of approaching the original problem. It is straight off our Differential Equations Sheet 1.

Let:

$f (x) = (\int_{0}^{x} e^{- t^{2}} dt)^{2}$

$g (x) = \int_{0}^{1} e^{- x^{2} (t^{2} + 1)} / (1 + t^{2}) dt$

Show that $f (x) + g (x) = π / 4$ . (Hint: differentiate the left hand side.)

By letting $x \to \infty$ , show that $\int_{0}^{\infty} e^{- t^{2}} dt = \sqrt{π / 4}$ .

This is equivalent (by symmetry) to the required result.

By Dave Sheridan (Dms22) on Thursday, March 29, 2001 - 01:26 pm :

Another way to view the integral-multiplied-by-an-integral is as follows.

Suppose c is a constant. Then the integral of cf(x) is the same as c times the integral of f(x). I think we're all pretty happy with that.

Well, what if c isn't a constant exactly but doesn't depend on x? For example, what's the integral of f(x)g(t) dx? Well, as far as x can see, t isn't anything to do with x so g(t) is a constant as x varies . So the integral is equal to g(t) times the integral of f(x)dx.

Finally, what's the integral of (the integral of g(t)dt)f(x)dx? Well, just as above we have that the integral of g(t)dt doesn't change if we change x, so it's a constant from the point of view of the integral dx. So we can conclude, just as above, that the integral is equal to:
(the integral of g(t)dt)(the integral of f(x)dx).

This argument allows us to swap things around easily, but is non-rigorous. Hopefully it makes sense though. Fubini's theorem states that we may interchange the order of integration, even if we can't factorise the integrals, so long as the integral itself is finite. To prove this properly requires a lot of complicated analysis, so I won't go into that here.

As for the confusion between multiplying two integrals together, the entity (integral of f(x)dx) is simply a number, so we can multiply integrals together as if we were simply multiplying two numbers.

-Dave

By Brad Rodgers (P1930) on Thursday, April 26, 2001 - 01:42 am :

Michael, I can't seem to figure how you can differentiate the LHS and end up with (presumably) 0. Upon differentiation, I get

4 e^{- x^{2}} \int_{0}^{x} e^{- t^{2}} dt

Have I missed something, or is there another method I need to use?

Thanks,

Brad

By Brad Rodgers (P1930) on Thursday, April 26, 2001 - 01:46 am :

Unless, of course, as I've just noticed, what one needs to do is let x tend to infinity after differentiation, then integrate. But is this allowed? If so, what allows us to do that?

Thanks,

Brad

By Kerwin Hui (Kwkh2) on Thursday, April 26, 2001 - 12:58 pm :

OK, let's do this step-by-step.

It is easy to see that

$f' (x) = \frac{d}{dx} {(\int_{0}^{x} \exp (- t^{2}) dt)}^{2} = 2 [\int_{0}^{x} \exp (- t^{2}) dt] \exp (- x^{2})$

$\begin{matrix} g' (x) & = & \int_{0}^{1} \frac{\partial}{\partial x} \frac{\exp [- x^{2} (t^{2} + 1)]}{1 + t^{2}} dt \\ = & - 2 x \int_{0}^{1} \exp [- x^{2} (t^{2} + 1)] dt = - 2 \exp (- x^{2}) \int_{0}^{x} \exp (- t'^{2}) dt' \end{matrix}$

and so $(f' + g') (x) \equiv 0$ for all $x$ .

Kerwin