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I am trying to partially prove that the characteristic funtion of a Gamma$(\alpha,\lambda)$ distribution, $\varphi(t)$, is such that $$ \varphi(t)=\left(\frac\lambda{\lambda-it}\right)^\alpha. $$ I am only trying to prove this for $|t|<\lambda$. However, my workings have led me to a different answer and I can't see what I have done wrong.

We have $$ \varphi(t)=\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}e^{itx}dx=\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\sum_{n=0}^\infty\frac{(itx)^n}{n!}dx. $$ Now I want to try to swap the integral and the sum, so I will try to apply Fubini's theorem. Consider $$ \begin{align*} \sum_{n=0}^\infty\int_0^\infty\left|\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}\right|dx=&\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(|t|x)^n}{n!}dx\\ =&\sum_{n=0}^\infty\left(\frac{|t|}\lambda\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ =&\sum_{n=0}^\infty\left(\frac{|t|}\lambda\right)^n\\ <&\infty \end{align*} $$ The third equality comes from comparing the integral to a Gamma$(\alpha+n,\lambda)$ density, and the fourth line (the inequality) comes from the fact that I assumed $|t|<\lambda$. So now we can swap the integral and the sum. So $$ \begin{align*} \varphi(t)&=\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}dx\\ &=\sum_{n=0}^\infty\left(\frac{it}{\lambda}\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ &=\sum_{n=0}^\infty\left(\frac{it}\lambda\right)^n\\ &=\frac\lambda{\lambda-it}. \end{align*}\\ $$ However, this is different to what I know the answer obviously should be, but I cannot see where I went wrong. Is this a valid approach? What should I do instead? Any help appreciated!

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Consider $$ \begin{align*} &\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(|t|x)^n}{n!}dx\\ =&\sum_{n=0}\frac{\Gamma(\alpha+n)}{\Gamma(\alpha)n!}\left(\frac{|t|}\lambda\right)^n\int_0^\infty\frac{\lambda^{\alpha+n}}{\Gamma(\alpha+n)}x^{\alpha+n-1}e^{-\lambda x}dx\\ =&\sum_{n=0}^\infty{\alpha+n-1\choose n}\left(\frac{|t|}\lambda\right)^n\\ =&\sum_{n=0}^\infty(-1)^n{-\alpha\choose n}\left(\frac{|t|}\lambda\right)^n\\ =&\sum_{n=0}^\infty{-\alpha\choose n}\left(-\frac{|t|}\lambda\right)^n\\ =&\left(1-\frac{|t|}\lambda\right)^{-\alpha}<\infty. \end{align*} $$ So we can swap the sum and the integral. And now we can apply a smiliar argument to get $$ \varphi(t)=\sum_{n=0}^\infty\int_0^\infty\frac{\lambda^\alpha}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x}\frac{(itx)^n}{n!}dx=\left(\frac{\lambda-it}\lambda\right)^{-\alpha}. $$ Edit: The only thing I am now unsure of is when did I use that $|t|<\lambda$?

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