Yet Another Sticky Integral

eep

227
0
I am now asked to integrate the following:

[tex]\int^\infty_{-\infty}xAe^{-\lambda(x-a)^2}dx = A\int^\infty_{-\infty}xe^{-\lambda(x-a)^2}dx[/tex]

Where A, a, and [itex]\lambda[/itex] are positive, real constants.

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

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

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

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

StatusX

Homework Helper
2,564
1
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

227
0
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

Homework Helper
2,564
1
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

Homework Helper
1,057
7
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

Homework Helper
1,057
7
eep said:
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. :smile:
 

StatusX

Homework Helper
2,564
1
By the way, in case you're interested (and since this seems to be the direction you're heading), indefinite integrals of the form:

[tex]\int x^n e^{-ax^2} dx[/tex]

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

[tex]\int_{-\infty}^{\infty} e^{-ax^2} dx=\sqrt{\frac{\pi}{a}}[/tex]

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

[tex]\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}}[/tex]

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):

[tex]\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[/tex]

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

eep

227
0
Thank you so much!
 

The Physics Forums Way

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top