How to Fix Limits of Integration for Gamma Integral #6?

[AhmedHesham]

Hi
I have a gamma integral in which it is not obvious how I can fix the limits of integration in order to match the standard form of gamma function.I just need someone to tell me how to fix them.

I mean the integral number 6 in the picture.
You can see my attempt in the PDF .

 

[anuttarasammyak]

$$I_6=\int_2^3 (3-y)^2 e^{-y^2}dy$$

$$I_6=\frac{1}{2}\int_4^9 (z^{1/2}-6+9z^{-1/2}) e^{-z}dz$$

By partial integration

$$\int_4^9 z^{1/2} e^{-z}dz=[-z^{1/2}e^{-z}]_4^9 + \frac{1}{2} \int_4^9 z^{-1/2}e^{-z}dz$$

Gauss error function works to estimate the integral of RHS.

 

[AhmedHesham]

$$I_6=\int_2^3 (3-y)^2 e^{-y^2}dy$$

$$I_6=\frac{1}{2}\int_4^9 (z^{1/2}-6+9z^{-1/2}) e^{-z}dz$$

By partial integration

$$\int_4^9 z^{1/2} e^{-z}dz=[-z^{1/2}e^{-z}]_4^9 + \frac{1}{2} \int_4^9 z^{-1/2}e^{-z}dz$$

Gauss error function works to estimate the integral of RHS.

Thanks for help
You didn't use the gamma function
The error function is not the subject of my course .
These exercises are on gamma function
Please make it in terms of gamma function.

 

[pasmith]

If you want to put the integrand in the form $f(t)e^{-t}$ then you are forced to take $t = (x - 3)^2$ and hence $$\int_0^1 x^2 e^{-(x-3)^2}\,dx = \int_4^9 \frac{(3 - \sqrt {t})^2}{2 \sqrt{t}}e^{-t}\,dt.$$ That's not a sum of gamma functions, because the limits are 4 and 9 not 0 and $\infty$. It can, however, be expresed as a sum of incomplete gamma functions.

 

[AhmedHesham]

Ok
I will ask my professor
Incomplete gamma function is beyond the scope of my course .
May be it was a mistake to give us it in the exercises on gamma function

 

[anuttarasammyak]

Please make it in terms of gamma function.

Incomplete gamma function and error function have the relation
$$\gamma(\frac{1}{2},x)=\sqrt{\pi}\ erf(\sqrt{x})$$

 

[AhmedHesham]

Incomplete gamma function and error function have the relation
$$\gamma(\frac{1}{2},x)=\sqrt{\pi}\ erf(\sqrt{x})$$

Ok
Thanks
You seem to be at a higher level.