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

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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.
Screenshot_2021-05-02-20-33-03-14_e2d5b3f32b79de1d45acd1fad96fbb0f.jpg

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

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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.
 
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anuttarasammyak said:
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.
 
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 <br /> \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.
 
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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
 
AhmedHesham said:
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})
 
anuttarasammyak said:
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.
 
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