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

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

Thanks,

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Welcome to MSE! If this is homework, it should be tagged as such. It really helps readability if you format your questions using MathJax. Did I get the edits correct? Also, what have you tried and where are you stuck? Regards –  Amzoti Feb 5 '13 at 19:06
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Let $w=\frac{a}{x}-u$. –  André Nicolas Feb 5 '13 at 19:12
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2 Answers

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

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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
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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.

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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
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