Two integrals I am trying to solve without closed form antiderivatives

In summary, to solve the integral of the functions x*exp(-a(x-b)^2) and x^2(exp(-a(x-b)^2), you can use integration by parts and substitution. The first term can be integrated using the error function and the second term can be integrated using substitution. The definite integral of the first term does not go to infinity if the calculations are done correctly.
  • #1
Ed Quanta
297
0
How do I solve the integral of the functions x*exp(-a(x-b)^2) and x^2(exp(-a(x-b)^2) where a and b are positive real numbers?

I tried integration by parts and cannot think of how to find the integral. In addition, while I was able to find exp(-a(x-b)^2) in an integral table, the two functions I listed above, I was not able to find. Any hints, suggestions, or integration tables you can point me towards would be appreciated.
 
Physics news on Phys.org
  • #2
Set [tex]u=\sqrt{a}(x-b)\to{dx}=\frac{du}{\sqrt{a}}[/tex]
Thus, for example, your first integrand transforms as:
[tex]\int{x}e^{-a(x-b)^{2}}dx=\frac{1}{a}\int(u+\sqrt{a}b)e^{-u^{2}}du=\frac{1}{a}\int{u}e^{-u^{2}}du+\frac{b}{\sqrt{a}}\int{e}^{-u^{2}}du[/tex]

the first term is easily integrated, the other is seen to be the error function integral that is well tabulated.
 
  • #3
Thanks.

I looked up the second term in an integration table (no problem). The first term I was able to integrate using substitution v = u^2 and dv=2udu. Here is my problem though. Can I solve this term as a definite integral with the bounds positive infinity and negative infinity? If so, how? I know the first term is going to equal -1/2a*exp(-v). But if I set the limit equal to negative infinity, this goes infinity. Does this term have a definite value?
 
  • #4
Nothing will go to infinity if you do the calculations correctly.
 

1. Can you explain what an integral is and why it is important?

An integral is a mathematical concept that represents the area under a curve on a graph. It is important because it allows us to calculate the total value or quantity of a changing quantity, such as velocity or volume, over a specific interval.

2. What is a closed form antiderivative and why is it useful in solving integrals?

A closed form antiderivative is an expression that can be written in terms of familiar mathematical functions, such as polynomials, logarithms, and trigonometric functions. It is useful in solving integrals because it provides a simpler and more straightforward way to evaluate the integral compared to using numerical methods.

3. What are some techniques for solving integrals without closed form antiderivatives?

Some techniques for solving integrals without closed form antiderivatives include using numerical methods, such as the trapezoidal rule or Simpson's rule, or using techniques such as integration by parts, substitution, or trigonometric identities.

4. How do you know if an integral has a closed form antiderivative?

There is no foolproof way to determine if an integral has a closed form antiderivative, but some common signs include the integrand containing only simple functions, such as polynomials, exponential functions, or trigonometric functions, and the limits of integration being constants or simple expressions.

5. Can you provide an example of an integral that does not have a closed form antiderivative?

Yes, the integral of the function e^x^2 does not have a closed form antiderivative. This type of integral is known as an "improper integral" and can only be evaluated using numerical methods or approximations.

Similar threads

Replies
8
Views
176
Replies
8
Views
1K
Replies
20
Views
2K
Replies
3
Views
1K
  • Calculus
Replies
6
Views
1K
Replies
3
Views
1K
Replies
8
Views
2K
Replies
3
Views
2K
  • Calculus
Replies
12
Views
2K
Replies
15
Views
2K
Back
Top