The Beta function is defined by the integral
$$B(\alpha,\beta)=\int_0^1x^{\alpha-1}(1-x)^{\beta-1}\,{\rm d}x~~~~(\operatorname{Re}\alpha,\operatorname{Re}\beta>0)$$
By evaluating $\int_0^\infty\int_0^\infty x^{\alpha-1}e^{-x}y^{\beta-1}e^{-y}\,{\rm d}x\,{\rm d}y$ in two different ways, show that
$$\Gamma(a)\Gamma(\beta)=\Gamma(\alpha+\beta)B(\alpha,\beta)$$
i have a proof of the relation between the gamma function and beta function but after you substitute the first time and swap the integrals why does the function become $x^{\alpha+\beta-1}$ after combinging $x^{\alpha-1}$ and $x^{\beta-1}$ shouldnt it be $x^{\alpha+\beta-2}$?
$$\begin{align*}
\Gamma(\alpha)\Gamma(\beta)&=\int_0^\infty x^{\color{blue}{\alpha-1}}e^{-x}\left(\int_0^\infty y^{\color{blue}{\beta-1}}e^{-y}\,{\rm d}y\right)\,{\rm d}x\\
&=\int_0^\infty x^{\color{blue}{\alpha+\beta-1}}e^{-x}\left(\int_0^\infty t^{\beta-1}e^{-tx}\,{\rm d}y\right)\,{\rm d}x&&(\text{put } y=tx)\\
&=\int_0^\infty t^{\beta-1}\left(\int_0^\infty x^{\alpha+\beta-1}e^{-(t+1)x}\,{\rm d}x\right)\,{\rm d}t\\
&=\int_0^\infty\frac{t^{\beta-1}}{(1+t)^{\alpha+\beta}}\left(\int_0^\infty u^{\alpha+\beta-1}e^{-u}\,{\rm d}u\right)\,{\rm d}t&&\left(\text{put }x=\frac u{1+t}\right)\\
&=\Gamma(\alpha+\beta)\int_0^\infty\frac{t^{\beta-1}}{(1+t)^{\alpha+\beta}}\,{\rm d}t
\end{align*}$$
Best Answer
Let's examine the crucial line in more detail. The substitution $y=tx$ gives $$\int_0^\infty y^{\beta-1}e^{-y}\,{\rm d}y\stackrel{y=tx}=\int_0^\infty(tx)^{\beta-1}e^{-tx}\color{red}{x}\,{\rm d}t=x^{\beta}\int_0^\infty t^{\beta-1}e^{-tx}\,{\rm d}t$$ As you can see, we have $x^{\beta-1}\cdot x=x^\beta$ from where the additional $-1$ disappears. That's all.