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Calculate the following integral

$$\int_0^{+\infty} \exp\left(-a^2 x\left(\dfrac{x-6}{x-2}\right)^2\right) \dfrac{dx}{\sqrt{x}}$$

I think the relation between this integral and function gamma is clear, But I do not know what variables change I do. Any Hint Pliss!

Yuriy S
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pablocn_
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    @ pablocn_ what about x=0 ?, it's seems not work in your domain – zeraoulia rafik Jun 29 '15 at 00:59
  • if its work in your domain it's seems to use inverse laplace transforma to error function – zeraoulia rafik Jun 29 '15 at 01:08
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    $I=\dfrac{\sqrt\pi}a.~$ An obvious first substitution is $t=\sqrt x~,$ since $\big(\sqrt x\big)'=\dfrac1{2\sqrt x}$ – Lucian Jun 29 '15 at 05:10
  • I was wondering if you considered applying residue calculus? http://mathfaculty.fullerton.edu/mathews/c2003/ResidueCalcMod.html – rtybase Jun 30 '15 at 22:40
  • @Lucian, can you provide some further hints? The first substitution is obvious, however I can't solve it further by any real methods I tried. The value is very surprising, since it's equal to just $$2 \int_0^{\infty} e^{-a^2 u^2} du$$ – Yuriy S Mar 21 '16 at 20:00
  • @YuriyS That is a Gaussian integral which can be done by https://en.wikipedia.org/wiki/Gaussian_integral – AHusain May 14 '16 at 10:26

1 Answers1

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Let $I$ be the integral in the question. First set $u^2=x$ so that: \begin{align} I=\int^\infty_0 \exp\left(-a^2x\left(1-\frac{4}{x-2} \right)^2\right)\,\frac{dx}{\sqrt[]{x}} =2\int^\infty_0 \exp\left(-a^2u^2\left(1-\frac{4}{u^2-2} \right)^2\right)\,du \end{align} Now notice that: $$I=\int^\infty_{-\infty} \exp\left(-a^2\left(u-\frac{4u}{u^2-2} \right)^2\right)\,du=\int^\infty_{-\infty} \exp\left(-a^2\left(u-\frac{2}{u+\sqrt[]{2}}-\frac{2}{u-\sqrt[]{2}} \right)^2\right)\,du$$ By Glasser's Master Theorem we get the result, namely: $$I=\int^\infty_{-\infty} \exp(-a^2u^2)\,du=\frac{\sqrt[]{\pi}}{|a|}$$ And we finally see the result Lucian suggested to be true three years ago.

Shashi
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