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How to prove $\int_{0}^{\infty}\left[\frac{\lambda}{2\pi x^3}\right]^{1/2}\exp\left\{\frac{-\lambda(x-\mu)^2}{2\mu^2 x}\right\}dx=1$?

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    I know I can do this. There is actually a [paper](http://amstat.tandfonline.com/doi/abs/10.1080/01621459.1968.10480942) to find the cdf. I just want to find out how to do the integral directly because of my mathematical curiosity.2012-12-06

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You can use the closed form formula for the more general integral

$\int_{ 0 }^{\infty} (ax^m)^{s} e^{\frac{-b(x-\mu)^2}{x}} = 2\,{a}^{s}{\mu}^{ms+1}{{\rm e}^{2b\mu}} {{\rm K_{m s+1}}\left(2\,b\mu\right)},$

where $\rm{K}_{\nu}(x)$ is the modified Bessel function of the second kind.

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Remember that google is your friend. Here are some sources where you can find the proof

Youtube:

http://www.youtube.com/watch?v=g_bCDcNWcgU

Math StackExchange:

How to show the normal density integrates to 1?

Wikipedia:

http://en.wikipedia.org/wiki/Gaussian_integral

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    These links seem to address a substantially different question of integrating a *Gaussian* density rather than the *inverse Gaussian*.2012-12-06