Mathematical Tools for Physics - Department of Physics - University

1—Basic Stuff 5

Ifnis a positive odd integer, these are elementary,

n= 1

∫∞

−∞

dxxne−αx

2

= 0 (nodd) (1.7)

To see why this is true, sketch graphs of the integrand for a few more oddn.

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

)

! (1.8)

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 case

n= 0andα= 1. Denote the integral byI, then

I=

∫∞

−∞

dxe−x

2

, and I^2 =

(∫∞

−∞

dxe−x

2

)(∫∞

−∞

dye−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

∫∞

−∞

dye−(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 becomesrdrdφ, 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

rdr

∫ 2 π

0

dφe−r

2

Theφintegral simply gives 2 π. For therintegral substituter^2 =z and the result is 1 / 2. [Or use

Eq. (1.8).] The two integrals together give youπ.

I^2 =π, so

∫∞

−∞

dxe−x

2

=

√

π (1.9)

Now do the rest of these integrals by parametric differentiation, introducing a parameter with

which to carry out the derivatives. Changee−x

2

toe−αx

2
, then in the resulting integral change variables
to reduce it to Eq. (1.9). You get

∫∞

−∞

dxe−αx

2

=

√

π

α

, so

∫∞

−∞

dxx^2 e−αx

2

=−

d

dα

√

π

α

=

1

2

(√

π

α^3 /^2

)

(1.10)

You can now get the results for all the higher even powers ofxby further differentiation with respect

toα.

* See section1.7in this chapter