1. ## integral

find

2. 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.

3. 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

4. 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

$\chi$ $\sigma$

5. 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