$\def\d{\mathrm{d}}$Prove that integration and summation can be swapped in the following expression:

$$\int_{0}^{\infty} \sum_{n=0}^{\infty} \frac{(itx)^n}{n!}f(x)\,\d x,$$

where $|t|<\lambda$ and $f(x)$ is the density function of the $\mathrm{Gamma}(\lambda, \alpha)$ distribution for any real $\alpha >0$.

I know that I need to use Fubini's theorem and show:

$$\int_{0}^{\infty} \sum_{n=0}^{\infty} \left|\frac{(itx)^n}{n!}f(x)\right|\,\d x< \infty,$$
or $$\sum_{n=0}^{\infty} \int_{0}^{\infty} \left|\frac{(itx)^n}{n!}f(x)\right|\,\d x< \infty.$$

I have tried the following: \begin{align*} &\mathrel{\phantom{=}} \sum_{n=0}^{\infty} \int_{0}^{\infty} \left|\frac{(itx)^n}{n!}f(x)\right|\,\d x\\ &< \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{\lambda^nx^n}{n!}|f(x)|\,\d x\ \text{(Used $|t|<\lambda$)}\\ &=\sum_{n=0}^{\infty} \int_{x=0}^{\infty} \frac{\lambda^nx^n}{n!}\left|\frac{x^{\alpha-1}\lambda^{\alpha}e^{-{\lambda}x}}{\Gamma(\alpha)}\right|\,\d x\ \text{(Substituted in $f(x)$)}\\ &=\sum_{n=0}^{\infty} \int_{x=0}^{\infty} \frac{1}{n!}\frac{x^{n+\alpha-1}\lambda^{n+\alpha}e^{-{\lambda}x}}{\Gamma(\alpha)}\,\d x\\ &\mathrel{\phantom{=}}\text{(Combined exponents and removed absolute value)}\\ &=\sum_{n=0}^{\infty} \frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)} \int_{x=0}^{\infty}\frac{x^{n+\alpha-1}\lambda^{n+\alpha}e^{-{\lambda}x}}{\Gamma(n+\alpha)}\,\d x\\ &=\sum_{n=0}^{\infty} \frac{\Gamma(n+\alpha)}{n!\Gamma(\alpha)}\\ &\mathrel{\phantom{=}}\text{(As the integral of the density of Gamma($\lambda$, $n+\alpha$) is $1$)} \end{align*}

The problem now is I don't believe this sum converges.

Any help would be appreciated!

  • 1
    $\begingroup$ Is your sum over $i $ or over $n $? Is the sum finite or infinite? $\endgroup$
    – PhoemueX
    Commented Apr 29, 2017 at 18:53
  • $\begingroup$ @PhoemueX The sum is over n and it is infinite. Thank you for spotting that typo $\endgroup$
    – rtyjkl3
    Commented Apr 30, 2017 at 8:03

1 Answer 1


You have two problems that prevent you from proving the claim:

  1. You effectively only use that $|t| \leq \lambda$, but you really do need to use $|t| < \lambda$.

  2. You correctly realized that you have to options to prove the claim ("I know that I need to use Fubini's theorem and show abc OR xyz"), but you tried to use the more difficult one.

First of all, note $$ \sum_{n=0}^\infty \left|\frac{(itx)^n}{n!}\right| = \sum_{n=0}^\infty (|t| \cdot x)^n / n! = e^{|t| \cdot x}. $$

Therefore, what we need to show is $$ \infty > \int_0^\infty e^{|t|\cdot x} \cdot f(x) dx. $$ From your attempt, I conclude that $f(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \cdot x^{\alpha - 1} \cdot e^{-\lambda x}$. Since the first factor is independent of $x$, it is not relevant for finiteness of the integral. Hence, we need to consider $$ \int_0^\infty e^{|t| \cdot x} \cdot x^{\alpha - 1} \cdot e^{\lambda x} dx = \int_0^1 x^{\alpha - 1} \cdot e^{(|t| - \lambda) x} \, dx + \int_1^\infty x^{\alpha - 1} \cdot e^{(|t| - \lambda) x} \, dx . $$ For the first integral, simply note that since $|t| \leq \lambda$ and since $x \geq 0$, we have $e^{(|t| - \lambda)x} \leq 1$, so that the first integral satisfies $$ \int_0^1 x^{\alpha - 1} \cdot e^{(|t| - \lambda) x} \, dx \leq \int_0^1 x^{\alpha - 1}\, dx = \frac{x^\alpha}{\alpha}\bigg|_0^1 = \frac{1}{\alpha}, $$ where we used that $\alpha > 0$.

Thus, it only remains to consider the second integral. To this end, let $\delta = \frac{\lambda - |t|}{2}$ and note $\delta > 0$. Now, for example from the Taylor expansion $e^x = \sum_n x^n / n!$, we have for $x \geq 1 > 0$ that $e^{\delta x} \geq (\delta x)^n / n! = C_{\delta, n} \cdot x^n \geq C_{\delta, n} \cdot x^{\alpha - 1}$ for arbitrary $n \in \Bbb{N}$ with $n \geq \alpha - 1$.

Therefore, $$ \int_1^\infty x^{\alpha - 1} e^{(|t| - \lambda)x} \, dx \leq \frac{1}{C_{\delta, n}} \cdot \int_1^\infty e^{\delta x} \cdot e^{-2\delta x} \, dx = \frac{1}{C_{\delta, n}} \cdot \int_1^\infty e^{-\delta x} \, dx < \infty, $$ since $\delta > 0$.


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