Mathematical Tools for Physics

1—Basic Stuff 7

To see why this is true, sketch a graph of the integrand (start with the casen= 1).
For the integral over positivexand still for oddn, do the substitutiont=αx^2.
∫∞

0

dxxne−αx

2

=

1

2 α(n+1)/^2

∫∞

0

dtt(n−1)/^2 e−t=

1

2 α(n+1)/^2

(

(n−1)/ 2

)

! (8)

Becausenis odd,(n−1)/ 2 is an integer and its factorial makes sense.
Ifnis even then doing this integral requires a special preliminary trick. Evaluate the special casen= 0
andα= 1. Denote the integral byI, then

I=

∫∞

−∞

dxe−x

2

, and I^2 =

(∫∞

−∞

dxe−x

2

)(∫∞

−∞

dy e−y

2

)

In squaring the integral you must use a different label for the integration variable in the second factor or it will
get confused with the variable in the first factor. Rearrange this and you have a conventional double integral.

I^2 =

∫∞

−∞

dx

∫∞

−∞

dy e−(x

(^2) +y (^2) )
This is something that you can recognize as an integral over the entirex-yplane. Now the trick is to switch to
polar coordinates*. The element of areadxdynow becomesr dr dθ, and the respective limits on these coordinates
are 0 to∞and 0 to 2 π. The exponent is justr^2 =x^2 +y^2.

I^2 =

∫∞

0

r dr

∫ 2 π

0

dθ e−r

2

Theθintegral simply gives 2 π. For therintegral substituter^2 =zand the result is 1 / 2. [Or use Eq. ( 8 ).] The
two integrals together give youπ.

I^2 =π, so

∫∞

−∞

dxe−x

2

=

√

π (9)

* See section1.7in this chapter