find

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- Mar 7th 2010, 02:11 AMdaporeintegral
find

- Mar 7th 2010, 08:57 AMshawsend
Write it as:

$\displaystyle \int x^{-3/2}-x^{-3/2} e^{-x} dx$

then use parts on the second one to get the expression $\displaystyle \frac{e^{-x}}{\sqrt{x}}$, then use the substitution $\displaystyle w=\sqrt{x}$ on that expression to get it in terms of $\displaystyle \int_0^{\infty}e^{-w^2}dw=\frac{\sqrt{\pi}}{2}$. May have to take limits since it's indeterminate at zero. Mathematica returns $\displaystyle 2\sqrt{\pi}$ but I haven't worked out all the steps so fix it if I left some kinks ok. - Mar 11th 2010, 01:24 PMdapore
- Mar 11th 2010, 02:27 PMchisigma
Integrating by parts we obtain...

$\displaystyle \int_{0}^{\infty} x^{-\frac{3}{2}}\cdot (1-e^{-x})\cdot dx = |-2\cdot x^{-\frac{1}{2}}\cdot (1-e^{-x})|_{0}^{\infty} +2\cdot \int_{0}^{\infty} x^{-\frac{1}{2}}\cdot e^{-x}\cdot dx = 2\cdot \Gamma(\frac{1}{2}) = 2\cdot \sqrt{\pi}$ (1)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Mar 11th 2010, 02:40 PMdapore