Ytukyg

I currently have, and have to calculate the Gamma function:

$$int_2^4 sqrt[4]{(x-2)(4-x)^3},mathbb{d}x$$

As per definition gamma function is:

$$int_0^1t^{z-1}e^{-t},mathbb{d}t$$

Do I understand correctly that first of all we have to reduce the lower and upper boundaries so they must be $0$ and $1$ respectivly?

To make lower boundory zero we have to substract $2$ and to make upper boundary 1 we have to also devide substitution by $2$ (as we have already had substrected $2$ and $2$ is left). As a result:

$$begin{array}{|c|} t = frac{x-2}{2} \ dt = frac{1}{2} dx \
x = 4t +2 end{array}$$

Substituting this into the integral will give us the Gamma function, and relationship between Gamma and Beta functions are clear, but I spent quite lot amount of time to understand what substitution is needed (and I am not sure if this one I used was right), so my question is:

What (if any) is the way to find out what substitution should be used in Euler's integral?

As a side note, the upper bound of your second integral should be $infty$ to get the $Gamma$ function.

A hint. Fubini's theorem is useful here. You can writebegin{align}
Gamma(x)Gamma(y) &= int_{0}^infty e^{-u} u^{x-1},du cdotint_{0}^infty e^{-v} v^{y-1},dv \[6pt]
&=int_{0}^inftyint_{0}^infty e^{-u-v} u^{x-1}v^{y-1},du ,dv,
end{align}
then by the change of variable $u=zt$, $v=z(1-t)$ one obtains
begin{align}
Gamma(x)Gamma(y) &= int_{0}^inftyint_{0}^1 e^{-z} (zt)^{x-1}(z(1-t))^{y-1}z,dt ,dz \[6pt]
&= int_{0}^infty e^{-z}z^{x+y-1} ,dzcdotbbox[5px,border:1px solid blue]{int_{0}^1t^{x-1}(1-t)^{y-1},dt}\
&=Gamma(x+y),{rm{B}}(x,y),
end{align}
as expected.