Evaluation of a definite integral

  • Thread starter dyn
  • Start date
  • #1
dyn
697
49
I am looking at a solution to an question and I don't understand how the value of the following definite integral comes out to be zero ? The following function is evaluated from 0 to ∞ with r being the variable

## exp(-β^2r^2)r^nr^-1/(n-1)## That should read r raised to the power of (n-1)
I presume when the value of ∞ is input the negative exponential turns out to be zero but I don't understand how inputting zero gives a value of zero ?
 
Last edited:

Answers and Replies

  • #2
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
I am looking at a solution to an question and I don't understand how the value of the following definite integral comes out to be zero ? The following function is evaluated from 0 to ∞ with r being the variable

## exp(-β^2r^2)r^nr^-1/(n-1)## That should read r raised to the power of (n-1)
which? where?
Did you mean to write: $$\int_0^\infty e^{-\beta^2r^2}r^{n-1}\;dr$$

I presume when the value of ∞ is input the negative exponential turns out to be zero but I don't understand how inputting zero gives a value of zero ?

What do you think it should come out to?
What is beta? What is "n"?
 
  • #3
dyn
697
49
which? where?
Did you mean to write: $$\int_0^\infty e^{-\beta^2r^2}r^{n-1}\;dr$$



What do you think it should come out to?
What is beta? What is "n"?

You have got the function right but it also has a factor of 1/(n-1). This function was derived from integration by parts. It just needs evaluating with the values ∞ and 0. When I input 0 I have exp(0) which is 1 and 0 to the power (n-1) which I thought is also 1. But overall the answer should be zero
 
  • #4
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
Oh you mean you've done an integral and you have ended up at this stage: $$\frac{1}{n-1}\left[ e^{-\beta^2r^2} r^{n-1}\right]_0^\infty$$
When I input 0 I have exp(0) which is 1 and 0 to the power (n-1) which I thought is also 1. But overall the answer should be zero
... when you put ##r=0## into ##r^{(n-1)}## you get ##0^{n-1}## right?
What is zero multiplied by itself? i.e. if n=3, then n-1=2 and ##0^2=0\times 0 = ?##
 
Last edited:
  • Like
Likes 1 person
  • #5
dyn
697
49
Thanks for that. Somehow I was under the impression that zero raised to any power was 1 !
 
  • #6
Simon Bridge
Science Advisor
Homework Helper
17,874
1,655
You just got a rule stuck in your head, that's all.
Happens to everyone ;)
 
  • #7
dyn
697
49
You just got a rule stuck in your head, that's all.
Happens to everyone ;)

Thanks. I did get the wrong rule stuck in my head. But what would happen at r=o if n=1 then I would have zero to the power of zero which I believe is 1 ? I also have a zero on the denominator. Is it impossible to evaluate in this case ? Incidentally the question did state n≥2 so I am just asking for future reference.
 
  • #8
SteamKing
Staff Emeritus
Science Advisor
Homework Helper
12,798
1,670
You should review the rules of exponentiation:

http://en.wikipedia.org/wiki/Exponentiation

The expression 0^0 (zero raised to the zeroth power) is considered, in general, an indeterminate form. 0^n is defined only for integers n > 0. There may be certain formulas which require that 0^0 be evaluated as equal to one, but these are exceptions.
 

Related Threads on Evaluation of a definite integral

  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
17
Views
1K
  • Last Post
Replies
7
Views
2K
Replies
11
Views
1K
  • Last Post
Replies
7
Views
5K
Replies
5
Views
1K
  • Last Post
Replies
5
Views
15K
  • Last Post
Replies
24
Views
2K
  • Last Post
Replies
3
Views
2K
Replies
4
Views
860
Top