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 write\begin{align} \Gamma(x)\Gamma(y) &= \int_{0}^\infty\ e^{-u} u^{x-1}\,du \cdot\int_{0}^\infty\ e^{-v} v^{y-1}\,dv \\[6pt] &=\int_{0}^\infty\int_{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}^\infty\int_{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} \,dz\cdot\bbox[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.