Efficient Integration of A, a, and \lambda in Sticky Integral: A Helpful Guide

[eep]

I am now asked to integrate the following:

$$\int^\infty_{-\infty}xAe^{-\lambda(x-a)^2}dx = A\int^\infty_{-\infty}xe^{-\lambda(x-a)^2}dx$$

Where A, a, and $\lambda$ are positive, real constants.

I first tried to do it by parts but that left me with infinities.

Then I tried setting $u = x - a$ and did some re-arranging, which left me with:

$$A\int^\infty_{-\infty}(u+a)e^{{-\lambda}u^2}dx = Aa\sqrt\frac{\pi}{\lambda}$$

Is this correct? Also, why does integrated by parts fail?

 

[StatusX]

Yea, that look's right. Integration by parts always works, it's just that sometimes it will give you the sum of two function which individually diverge on the interval you're interested in (maybe one to positive infinity and one to negative infinity), but are such that the limit of their sum is finite.

 

[eep]

Ah, I see. I followed the method you posted last time to solve only the exponential part of the integral, but what method can be used to solve this? I had to look up the answer in a table, which I don't mind doing, but I feel it's important to do the integral by hand at least once.

 

[StatusX]

The lefthand side of your last equation is a sum of two functions, one which is odd and integrated over an even range, so it integrates to zero, and one which you did in the other post.

 

[benorin]

In "Principles of Mathematical Analysis" by Rudin, ch. 6, exercise #9 indicates that: There are hypotheses required in order that integration by parts may be applied to improper integrals (the ones with infinite bounds of integration,) but it doesn't specify what thoses hypotheses are (that's the exercise.)

 

[benorin]

I had to look up the answer in a table, which I don't mind doing, but I feel it's important to do the integral by hand at least once.

This is wonderful. Thank you.

 

[StatusX]

By the way, in case you're interested (and since this seems to be the direction you're heading), indefinite integrals of the form:

$$\int x^n e^{-ax^2} dx$$

do not have simple closed forms for n even. But there is a trick if your bounds are infinite. Starting with:

$$\int_{-\infty}^{\infty} e^{-ax^2} dx=\sqrt{\frac{\pi}{a}}$$

You can repeatedly differentiate both sides with respect to a to get:

$$\frac{\partial}{\partial a} \left( \int_{-\infty}^{\infty} e^{-ax^2} dx \right) = \int_{-\infty}^{\infty} (-x^2) e^{-ax^2} dx =\frac{-1}{2}\sqrt{\frac{\pi}{a^3}}$$

and so on. You can also integrate from zero to infinity by noting that these functions are even. For n odd, you can get an explicit formula for the antiderivative. For simplicity, assume a=1 (you can figure out what to do otherwise). Then use the substitution u=x^2 (assume n=2k+1):

$$\int x^n e^{-x^2} dx=\frac{1}{2}\int x^{2k} e^{-x^2} ( 2xdx)=\frac{1}{2}\int u^k e^{-u} du$$

Which can be integrated by parts (or by looking up the gamma function if your bounds are again infinite).

 

[eep]

Thank you so much!