Evaluation of a definite integral

In summary, when evaluating the definite integral $$\int_0^\infty e^{-\beta^2r^2}r^{n-1}\;dr$$, the value of 0^0 should not be considered as 1. The expression is an indeterminate form and should not be evaluated.
  • #1
dyn
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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 ?
 
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  • #2
dyn said:
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
Simon Bridge said:
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
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 = ?##
 
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  • #5
Thanks for that. Somehow I was under the impression that zero raised to any power was 1 !
 
  • #6
You just got a rule stuck in your head, that's all.
Happens to everyone ;)
 
  • #7
Simon Bridge said:
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
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.
 

1. What is a definite integral?

A definite integral is a mathematical concept used to find the area under a curve between two specific points on a graph. It is represented by the symbol ∫ and has both a lower and upper limit of integration.

2. How is a definite integral evaluated?

A definite integral can be evaluated using various methods such as the Riemann sum, the Trapezoidal rule, or the Simpson's rule. These methods involve breaking down the area under the curve into smaller, simpler shapes and calculating their individual areas. The sum of these areas gives an approximation of the definite integral, which becomes more accurate as the number of smaller shapes increases.

3. What is the importance of evaluating a definite integral?

Evaluating a definite integral is important in many fields of science and engineering, as it allows us to find the total value or quantity of a changing quantity over a specific interval. This can be used to solve real-world problems related to areas, volumes, and rates of change.

4. Can definite integrals have negative values?

Yes, definite integrals can have negative values. This occurs when the curve lies below the x-axis, resulting in a negative area. The negative value of a definite integral represents the signed area under the curve, taking into account both positive and negative areas.

5. Are there any practical applications of definite integrals?

Yes, definite integrals have numerous practical applications in fields such as physics, economics, engineering, and statistics. They are used to calculate quantities such as work, displacement, velocity, and acceleration in physics, as well as to find the total cost and revenue in economics. In engineering, definite integrals are used to determine the volume and surface area of complex shapes. They are also used in statistics to find the probability of events occurring within a specific interval.

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