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This is my third question following the previous post.

Prove that \begin{equation} \int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^\infty x^{\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx \end{equation}

I am really having trouble to prove it. Please help me to compute the integrals. Every answer would be greatly appreciated. Thank you.

Anastasiya-Romanova 秀
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1 Answers1

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Examining the original integral, use the substitution $u=x^{-1}$ so that $$\frac{du}{dx}=-x^{-2}\Rightarrow dx=-(u^{-2})du$$ and $u\rightarrow 0$ when $x\rightarrow\infty$, $u\rightarrow \infty$ when $x\rightarrow0$

Therefore $$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=-\int_{\infty}^0u^{\frac{3}{2}}e^{-\frac{(1-u)^2}{u}}(u^{-2})du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(1-u)^2}{u}}du\\=\int_{0}^{\infty}u^{-\frac{1}{2}}e^{-\frac{(u-1)^2}{u}}du$$

If we replace the variable $u$ by $x$ (which is equivalent to the trivial substitution $x=u$) we will end up with the desired result:- $$\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$

Next we apply integration by parts to the integral in the RHS of the above equation, by setting $dv=x^{-\frac{1}{2}}$ and $u=e^{-\frac{(x-1)^2}{x}}$ so that $v=2x^{\frac{1}{2}}$ and $du=e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)$ and recalling that $$\int_0^{\infty}(u)(dv)dx=[uv]_0^\infty-\int_0^{\infty}v(du)dx$$ leading to $$\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx=\left[2x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\right]_0^{\infty}-2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}\left(\frac{1}{x^2}-1\right)dx\\=0-2\int_0^{\infty}x^{-\frac{3}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\=-2\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx+2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx \\\Rightarrow 3\int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=2\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx\\\Rightarrow \int_0^{\infty}x^{-\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx=\frac{2}{3}\int_0^{\infty}x^{\frac{1}{2}}e^{-\frac{(x-1)^2}{x}}dx$$

Let $I_n$ denote the integral we wish to evaluate. Making use of the equality of the integrals in the first part of the question, we have $$2I_n=\int_0^\infty x^{-\large\frac{1}{2}}e^{-\large\frac{(x-1)^2}{x}}dx+\int_0^\infty x^{-\large\frac{3}{2}}e^{-\large\frac{(x-1)^2}{x}}dx$$ Using the substitution $u=x^{\frac{1}{2}}$ we have $\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$ , resulting in $$2I_n=2\int_0^\infty e^{-\large\left(u-\frac{1}{u}\right)^2}du+2\int_0^\infty\frac{1}{u^2} e^{-\large\left(u-\frac{1}{u}\right)^2}du \\\Rightarrow I_n=\int_0^\infty \left(1+\frac{1}{u^2}\right) e^{-\large\left(u-\frac{1}{u}\right)^2}du$$ If we use the substitution $s=u-\frac{1}{u}$ , we have $ds=du\left(1+\frac{1}{u^2}\right)$ so that the integral reduces to a Gaussian integral (note the change in limits, as $s\rightarrow-\infty$ when $u\rightarrow 0$ and $s\rightarrow\infty$ when $u\rightarrow \infty$):- $$I_n=\int_{-\infty}^\infty e^{-s^2}ds=\sqrt{\pi}$$

Alijah Ahmed
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  • Wow! Cool! Thank you so much Mr. Alijah Ahmed. I really appreciate it your answer since no-one wants to answer this question. Using Wolfram I get the integrals equal $\sqrt{\pi}$. How I get the answer equals $\sqrt{\pi}$? – Anastasiya-Romanova 秀 May 11 '14 at 14:11
  • I really hope you will evaluate the integral for me but never mind, I still accept your answer, though. (─‿‿─) – Anastasiya-Romanova 秀 May 18 '14 at 11:26
  • @V-Moy - well done on proving the integral, and thank you for the link you have posted. Although I guess I am too late, I have filled in the answer anyway for completeness - I was trying to get to the Gaussian integral form. – Alijah Ahmed May 19 '14 at 19:06
  • It's okay, thanks for your help (ô‿ô) – Anastasiya-Romanova 秀 May 20 '14 at 04:26