Evaluate # int_(-oo)^(oo) \ x^2e^(-4x) \ dx# ?
1 Answer
Nov 2, 2017
The integral is divergent.
Explanation:
We seek:
# int_(-oo)^(oo) \ x^2e^(-4x) \ dx #
Using common sense analysis we can answer the question without actually performing the integral.
If we apply Integration By Parts twice we will get a solution of the definite integral of the form:
# int \ x^2e^(-4x) \ dx = (Ax^2+Bx+C)e^(-4x) + C #
Thuis we will find that:
# int_(-oo)^(oo) \ x^2e^(-4x) \ dx = lim_(N rarr oo)[(Ax^2+Bx+C)e^(-4x)]_(-N)^(N)#
And will will get convergence at the upper limit as
Hence the integral is divergent.
Note:
Two applications of IBP yields:
# int \ x^2e^(-4x) \ dx = -((8x^2+4x+1)e^(-4x))/32 + C#