integral

**dapore** · March 7th 2010, 02:11 AM

find

**shawsend** · March 7th 2010, 08:57 AM

Write it as:

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

then use parts on the second one to get the expression $\frac{e^{-x}}{\sqrt{x}}$ , then use the substitution $w=\sqrt{x}$ on that expression to get it in terms of $\int_0^{\infty}e^{-w^2}dw=\frac{\sqrt{\pi}}{2}$ . May have to take limits since it's indeterminate at zero. Mathematica returns $2\sqrt{\pi}$ but I haven't worked out all the steps so fix it if I left some kinks ok.

**dapore** · March 11th 2010, 01:24 PM

Originally Posted by shawsend

Write it as:

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

then use parts on the second one to get the expression $\frac{e^{-x}}{\sqrt{x}}$ , then use the substitution $w=\sqrt{x}$ on that expression to get it in terms of $\int_0^{\infty}e^{-w^2}dw=\frac{\sqrt{\pi}}{2}$ . May have to take limits since it's indeterminate at zero. Mathematica returns $2\sqrt{\pi}$ but I haven't worked out all the steps so fix it if I left some kinks ok.

Thank you, but my teacher says this method is not working, and says that the correct way is to insert (x)in the bow and the integration is a two parts, one directly and the other gamma function

**chisigma** · March 11th 2010, 02:27 PM

Integrating by parts we obtain...

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

Kind regards

$\chi$ $\sigma$

**dapore** · March 11th 2010, 02:40 PM

Originally Posted by chisigma

Integrating by parts we obtain...

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

Kind regards

$\chi$ $\sigma$

Thank you, my dear, indeed this is the answer

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