# Thread: Why does it converge?

1. ## Why does it converge?

Hi, I was wondering if someone could help me out in understanding the result in the image I'm attaching below. I'm OK with the entire derivation, except for the very first line - the author simply asserts that "The improper integral converges [...]" and I'm feeling a bit lost at how exactly we know that it does converge. Any pointers would be great!

2. ## Re: Why does it converge?

Consider the double integral \displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) }\,\mathrm{d}x}\,\mathrm{d}y} \end{align*}. This can be evaluated by converting to polars. Since we are integrating over the entire real line, that means in polars \displaystyle \begin{align*} 0 \leq r \leq \infty \end{align*} and \displaystyle \begin{align*} 0 \leq \theta \leq 2\pi \end{align*}. We have

\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) } \, \mathrm{d}x}\,\mathrm{d}y} &= \int_0^{2\pi}{\int_0^{\infty}{\mathrm{e}^{-r^2}\,r\,\mathrm{d}r}\,\mathrm{d}\theta} \\ &= -\frac{1}{2} \int_0^{2\pi}{\int_0^{\infty}{-2r\,\mathrm{e}^{-r^2} \,\mathrm{d}r}\,\mathrm{d}\theta} \\ &= -\frac{1}{2} \int_0^{2\pi}{\int_0^{-\infty}{\mathrm{e}^u\,\mathrm{d}u}\, \mathrm{d} \theta} \textrm{ after substituting } u = -r^2 \implies \mathrm{d}u = -2r\,\mathrm{d}r \\ &= \frac{1}{2} \int_0^{2\pi}{\int_{-\infty}^0{\mathrm{e}^u\,\mathrm{d}u }\,\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}\int_{\epsilon}^0{\mathrm{e}^u\,\mathrm{d}u }\,\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}{ \left[ \mathrm{e}^u \right] _{\epsilon}^0 }\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}{ \left( \mathrm{e}^0 - \mathrm{e}^{\epsilon} \right) }\,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\pi}{ \left( 1 - 0 \right) \, \mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\pi}{1\,\mathrm{d}\theta} \\ &= \frac{1}{2} \left[ \theta \right] _0^{2\pi} \\ &= \frac{1}{2} \left( 2\pi - 0 \right) \\ &= \pi \end{align*}

Now let's try to evaluate the double integral in another way...

\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) } \,\mathrm{d}x}\,\mathrm{d}y} &= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{e}^{-y^2}\, \mathrm{d}x}\,\mathrm{d}y} \\ &= \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} \end{align*}

Notice that these two integrals are numerically identical, and thus \displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} \end{align*}, giving

\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} &= \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x}\int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \\ &= \left( \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \right) ^2 \end{align*}

But we already know what this integral's value is, it's \displaystyle \begin{align*} \pi \end{align*}, thus

\displaystyle \begin{align*} \left( \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \right) ^2 &= \pi \\ \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} &= \sqrt{\pi} \end{align*}

3. ## Re: Why does it converge?

Thanks for the reply - however, I never had any problem understanding the derivation AFTER the assumption that the integral converges; it's the author's assertion that "The improper integral converges [...]" that's leaving me in the dark. How can we simply look at the original form of the integral and tell right away that it does indeed converge?

4. ## Re: Why does it converge?

If you can show that an integral does in fact equal a number, then it is convergent. That is the very definition of convergence...

You can probably get a decent picture before trying to evaluate the integral if you realise that the function is even, and that the negative exponential decays at a very fast rate.

5. ## Re: Why does it converge?

Prove It showed that it converges when he wrote (I have dropped the second integral- as it is not improper, it doesn't affect the convergence):

$\int_{-\infty}^0 e^{-u}du= \lim_{\epsilon\to -\infty}\int_{\epsilon}^0 e^{u}du= \lim_{\epsilon\to -\infty} \left[e^{-u}\right]_{\epsilon}^0= \lim_{\epsilon\to -\infty} (e^0- e^{\epsilon})= 1- 0$

You show that an improper integral converges by showing that the limit exists. Here the point is that $\lim_{\epsilon\to -\infty} e^{\epsilon}= 0$.
Since that limit exists, the improper integral converges.

6. ## Re: Why does it converge?

Here's another way you might want to look at it. Suppose that

$0 \le f(x) \le g(x)$

and

$\int_0^{\infty} g(x) dx$ converges then $\int_0^{\infty} f(x) dx$ converges.

I don't think it's difficult to show that

$e^{-x^2} \le e^{1-x}$

so that

$\int_0^{\infty}e^{-x^2}dx \le\int_0^{\infty} e^{1-x}dx = e.$