Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I want to ask how we can integrate the following function, $f(T)$, which is near Gaussian.

$$ \large \intop_{-k}^k \frac{2a}{x^{2}} \cdot {e^ {-(\frac{a}{x}-u)^{2}/b^2}} dx$$


share|cite|improve this question
Let $w=\frac{a}{x}-u$. – André Nicolas Feb 5 '13 at 19:12

Hint: Let $v=\frac{a}{x}-u$, then $dv=\frac{-a}{x^2}$: $$\large \intop_{\frac{a}{-k}-u}^{\frac{a}{k}-u} -{} 2{e^ {-(v)^{2}/b^2}} dv$$

share|cite|improve this answer
Thanks man, that was helpful. I worked it out to the definite integral and use the error function. – user61108 Feb 6 '13 at 0:47

Hint: Let $w=\dfrac{a}{x}-u$. Then $dw=-\dfrac{a}{x^2}\,dx$. That will bring you to familiar territory.

Or else you can make a trickier substitution and reduce to the standard normal in one step.

share|cite|improve this answer
Thanks for your response. I just clarify the function again. do u think the integration by parts will work with your hint? – user61108 Feb 5 '13 at 19:18
@user61108: The hint reduces the problem to the integral of a constant times $e^{-w^2/b^2}$. This does not have an elementary antiderivative. With the right substitution, you can reduce it to a definite integral of $e^{-t^2/2}$, in other words of the standard normal density function. One cannot get a formula for the indefinite integral in terms of elementary functions. – André Nicolas Feb 5 '13 at 19:24
I tried to substitute t=2w/b to get e^-t^2/2 but get w^2 now in the integral. any advice? – user61108 Feb 5 '13 at 19:40
Yes, $t=\sqrt{2}w/b$, or equivalently $w=tb/\sqrt{2}$. – André Nicolas Feb 5 '13 at 19:42
thank you again. now i got dt and reduce to the definite integral. Now i can use erf with the updated limits. Am I right? – user61108 Feb 5 '13 at 19:57

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.