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all. I am trying integrate this equation(gamma density). $\int\limits_0^\infty \frac{1}{\sqrt{\left| x-1 \right|}}\exp \left( -\left| 1-x \right| \right) \;dx$

What I have done is split it into 2 cases, but stuck on integrating with square root.

$f(x)= \begin{cases} \int\limits_{0}^{1}{\frac{1}{\sqrt{1-x}}\exp \left( -(1-x) \right)}\;dx & x<1 \\ \int\limits_{1}^{\infty }{\frac{1}{\sqrt{x-1}}\exp \left( -(x-1) \right)}\;dx & x>1 \\ \end{cases}$

How do I integrate each one? Thanks!

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    thanks for retag Kannappan. Michael, I wasn't trying to say things in the simplest way possible, but I agree with your point. André, can you elaborate? Wouldn't this be applicable to first equation?2012-02-13

2 Answers 2

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In the former integral, the substitution $u=\sqrt{1-x}$ gives

$\int_1^0 e^{-u^2}(-2du)=\sqrt{\pi}\operatorname{erf}(1). $

In the latter integral, the substitution $u=\sqrt{x-1}$ gives

$\int_0^\infty e^{-u^2}(2du)=\sqrt{\pi}.$

Put them together and we have $\sqrt{\pi}\big(1+\operatorname{erf}(1)\big)$. Here $\operatorname{erf}$ is the error function.

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    I was not being sarcastic. The negative sign in the equation comes from $du = \color{Red}-\frac{1}{2}dt/u$. The substitution switches the end points to $\int_1^0$, which by the fundamental theorem we can switch back to $\color{Blue}- \int_0^1$. These two negative signs cancel each other out.2012-02-14
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The second one is easy (with $x-1=t^2$) : $\int_0^{\infty} \frac1{\sqrt{t^2}} e^{-t^2} 2t\ dt=2\int_0^{\infty}e^{-t^2}dt=\sqrt{\pi}$

The first one (with $1-x=t^2$) gives : $-\int_1^0 \frac1{\sqrt{t^2}} e^{-t^2} 2t\ dt=2\int_0^1 e^{-t^2}dt= \sqrt{\pi} \ \mathrm{erf}(1)$ by definition of the error function

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    wait, the bounds you give are not the same as the question.2012-02-14