If ${X_1} \sim\gamma({\alpha _1},{\beta _1})$ and ${X_2} \sim\gamma({\alpha _2},{\beta _2})$, I need to prove ${X_1}+{X_2}∼Γ({\alpha _1}+{\alpha _1},{\beta _1}*{\beta _2})$ if ${X_1}$ and ${X_2}$ are independent.
I know that if ${X_1}$ and ${X_2}$ are independent continuous random variables. The probability density function of $y={X_1}+{X_2}$ is given by the convolution of the probability density functions, This is what I have done so far: $$ \begin{align} f_{x_1+x_2}(y)&=\int_{-\infty}^\infty f_{x_1}(x)f_{x_2}(y - x)dx\\&=\int_{-\infty}^\infty\left(\frac{1}{\beta^{\alpha_1}_1}\Gamma(\alpha_1)x^{\alpha_1-1}e^{-\frac{x}{\beta_1}} \right)\left(\frac{1}{\beta^{\alpha_2}_2}\Gamma(\alpha_2)(y - x)^{\alpha_2-1}e^{-\frac{(y-x)}{\beta_2}}\right)dx\end{align} $$ note: $ \begin{array}{c} x \in [0,\infty )\\ y - x \in [0,\infty )\\ y - x \ge 0 \to y \ge x \end{array}$ then $$ = \frac{1}{{\beta _1^{{\alpha _1}}\Gamma {\alpha _1}\beta _2^{{\alpha _2}}\Gamma {\alpha _2}}}\int\limits_0^y {{x^{{\alpha _1} - 1}}(} y - x{)^{{\alpha _2} - 1}}{e^{ - \frac{x}{{{\beta _1}}}}}{e^{ - \frac{{(y - x)}}{{{\beta _2}}}}}dx $$ let $ u = \frac{x}{y} \Rightarrow du = \frac{{dx}}{y} \Rightarrow dx = ydu $
so $$ = \frac{1}{{\beta _1^{{\alpha _1}}\Gamma {\alpha _1}\beta _2^{{\alpha _2}}\Gamma {\alpha _2}}}\int\limits_0^1 {{x^{{\alpha _1} - 1}}(} y - x{)^{{\alpha _2} - 1}}{e^{ - \frac{x}{{{\beta _1}}}}}{e^{ - \frac{{(y - x)}}{{{\beta _2}}}}}dx $$ $$ = \frac{1}{{\Gamma {\alpha _1}\Gamma {\alpha _2}\beta _1^{{\alpha _1}}\beta _2^{{\alpha _2}}}}\int\limits_0^1 {{{(yu)}^{{\alpha _1} - 1}}(} y - yu{)^{{\alpha _2} - 1}}{e^{ - \frac{{uy}}{{{\beta _1}}}}}{e^{ - \frac{{(y - yu)}}{{{\beta _2}}}}}ydu $$ $$ = \frac{1}{{\Gamma {\alpha _1}\Gamma {\alpha _2}\beta _1^{{\alpha _1}}\beta _2^{{\alpha _2}}}}\int\limits_0^1 {{u^{{\alpha _1} - 1}}{y^{{\alpha _1} - 1}}{y^{{\alpha _2} - 1}}(} 1 - u{)^{{\alpha _2} - 1}}{e^{ - \frac{{uy}}{{{\beta _1}}}}}{e^{ - \frac{{(y - yu)}}{{{\beta _2}}}}}ydu $$ $$ = \frac{1}{{\Gamma {\alpha _1}\Gamma {\alpha _2}\beta _1^{{\alpha _1}}\beta _2^{{\alpha _2}}}}\int\limits_0^1 {{u^{{\alpha _1} - 1}}{y^{{\alpha _1} + {\alpha _2} - 1}}(} 1 - u{)^{{\alpha _2} - 1}}{e^{ - \frac{{uy}}{{{\beta _1}}}}}{e^{ - \frac{{(y - yu)}}{{{\beta _2}}}}}du $$ $$ = \frac{{{y^{{\alpha _1} + {\alpha _2} - 1}}}}{{\Gamma {\alpha _1}\Gamma {\alpha _2}\beta _1^{{\alpha _1}}\beta _2^{{\alpha _2}}}}\int\limits_0^1 {{u^{{\alpha _1} - 1}}(} 1 - u{)^{{\alpha _2} - 1}}{e^{ - \frac{{uy}}{{{\beta _1}}}}}{e^{ - \frac{{(y - yu)}}{{{\beta _2}}}}}du $$ But I'm stuck here, I don't know how to solve this term $$ {e^{ - \frac{{uy}}{{{\beta _1}}}}}{e^{ - \frac{{(y - yu)}}{{{\beta _2}}}}} $$ Can someone help me? Thanks in advance