# Prove this integral

1. Dec 20, 2005

### maverick6664

Hi all,
Reading a book on quantum mechanics, I cannot understand an integral.
Now I can understand this one.
however I don't understand this.
As a matter of course, x, t are real numbers.
Maybe I'm missing something easy....
Thanks in advance! (btw how can I display these as images? I'm using img tags)

Last edited by a moderator: Apr 21, 2017
2. Dec 20, 2005

### matt grime

substitution x=t+i and not worrying too much seems to make it make more sense.

3. Dec 20, 2005

### arildno

Consider the contour C in the complex plane consisting of the line segments (separately parametrized with "t"):
$L_{1}: t+i, -a\leq{t}\leq{a}$
$L_{2}: a+(1-t)i, 0\leq{t}\leq{1}$
$L_{3}: -t, -a\leq{t}\leq{a}$
$L_{4}: a+ti 0\leq{t}\leq{1}$

Consider now the (complex) integral $\oint_{C}exp(-z*z)dz$
where the complex variable z is to be evaluated along the contour C.
Since the integrand is an analytical function we have, by Cauchy's theorem:
$\oint_{C}exp(-z*z)dz=0$
This is analogous to the real (multi-)variable theorem that says that the integral of a gradient field along a closed contour is zero.

Furthermore, w the complex integral is additive, so we may split up the integral over C in 4 integrals over the 4 line segments:
$\oint_{L_{1}}exp(-z*z)dz+\oint_{L_{2}}exp(-z*z)dz+\oint_{L_{3}}exp(-z*z)dz+\oint_{L_{4}}exp(-z*z)dz=0$

Now, let us look at the limiting expression when we let "a" go towards infinity:
Every complex point on the vertical strips $L_{2},L{4}[/tex] will get bigger and bigger modulus. But that means that the two integrals along these strips will decrease in value, reaching 0 in the limit. Thus, we are left with the expression: [itex]\int_{-\infty}exp(-(t+i)*(t+i))dt+\int_{\infty}exp(-t*t)dt=0$
Here, the upper limit in the first integral is infinity, whereas the upper limit in the second integral is negative infinity. (My keyboard is working against me!)

Switching upper and lower limits in the last integral effects the identity you were after.

4. Dec 20, 2005

### maverick6664

Thanks! It's so easy...

I noticed

exp(-(t+i)^2) is the conjugate of exp(-(-t+i)^2) !!

arildno's explanation is very interesting.

Last edited: Dec 20, 2005
5. Dec 21, 2005

### maverick6664

Let me practice tex on this forum:
$$exp(-(t+i)^2)$$
is the conjugate of
$$exp(-(-t+i)^2)$$

So

$$\int^{+\infty}_{-\infty} exp(-(x+i)^2)dx = \int^{+\infty}_{_\infty} exp(-x^2)dx$$

Oh! cool tex implementation!!

Last edited: Dec 21, 2005