- #1
foo_daemon
- 8
- 0
Hi,
I need to solve the following integral from 0 to [tex]\infty[/tex] :
Please note that my professor has defined [tex] log(x) = ln(x) [/tex] , i.e. 10 is not the default base.
[tex]\int { e^{\frac{-log(x)^2}{2} } dx }[/tex]
Through 'simplification' ( [tex] e^{log(x)} = x [/tex] ), I have translated the function to:
[tex] \int { x^{ \frac{-log(x)}{2} } dx } [/tex] , which appears easier to integrate.
However, I am uncertain where to go from here (or if this is even the right direction). Obviously the formula [tex] \int { x^c } = \frac{x^{c+1}}{c+1}[/tex] doesn't work here, as c is not a constant.
I have tried using mathematica to see if I could reverse-engineer the integral, but the resulting 'error' function seems significantly complex and doesn't help.
I have tried substitution with it in the form [tex] \frac{ 1}{x^{\frac{log(x)}{2}} } [/tex] using [tex] u = \frac{log(x)}{2} \ \ du = \frac {1}{2x} dx [/tex] , which almost works, except that the problem then becomes [tex] \int {2 du^u } [/tex] , which seems quite odd.
Any hints for how to approach this problem?
I need to solve the following integral from 0 to [tex]\infty[/tex] :
Please note that my professor has defined [tex] log(x) = ln(x) [/tex] , i.e. 10 is not the default base.
[tex]\int { e^{\frac{-log(x)^2}{2} } dx }[/tex]
Through 'simplification' ( [tex] e^{log(x)} = x [/tex] ), I have translated the function to:
[tex] \int { x^{ \frac{-log(x)}{2} } dx } [/tex] , which appears easier to integrate.
However, I am uncertain where to go from here (or if this is even the right direction). Obviously the formula [tex] \int { x^c } = \frac{x^{c+1}}{c+1}[/tex] doesn't work here, as c is not a constant.
I have tried using mathematica to see if I could reverse-engineer the integral, but the resulting 'error' function seems significantly complex and doesn't help.
I have tried substitution with it in the form [tex] \frac{ 1}{x^{\frac{log(x)}{2}} } [/tex] using [tex] u = \frac{log(x)}{2} \ \ du = \frac {1}{2x} dx [/tex] , which almost works, except that the problem then becomes [tex] \int {2 du^u } [/tex] , which seems quite odd.
Any hints for how to approach this problem?
Last edited: