Switching the Order of Summation and Integration: Incomplete Gamma Function

Question

I want to solve the following integral:

$$ I(x) = \int_0^\infty \gamma(N, x) f(x) \, dx $$

where $|f(x)| \leq 1$ and $\gamma(N, x)$ is the incomplete Gamma function with $N \in \mathbb{Z}_{++}$. To do so, I expand the incomplete Gamma function:

$$ I(x) = \Gamma(N) \int_{0}^{\infty} x^N e^{-x} \sum_{k=0}^\infty \frac{x^k}{\Gamma(N + k + 1)} f(x) \, dx $$

Now, I want to switch the order of integration and summation. The question is: Can I do so?

According to: https://en.wikipedia.org/wiki/Incomplete_gamma_function. The sum is locally uniformly convergent. Given that the summmands are clearly continuous, is local uniform convergence enough to allow me to switch these operations?

---

Thanks!

Answer 1

Let $$ \int_{0}^{\infty }f(x)\left(\sum_{n\geqslant 0}g_n(x)\right)\,\mathrm d x\tag1 $$

where $|f(x)|\leqslant 1$ and $g_n\geqslant 0$ for all $n\in \Bbb N $. Then $$ \int_{0}^{\infty }f(x)\left(\sum_{n\geqslant 0}g_n(x)\right)\,\mathrm d x\\ \qquad=\sum_{n\geqslant 0}\int_{0}^\infty f^+(x)g_n(x)\,\mathrm d x- \sum_{n\geqslant 0}\int_{0}^\infty f(x)^-g_n(x)\,\mathrm d x\tag2 $$ if the RHS is well-defined (that is, if at least one of the integrals is finite), where $f(x)=f(x)^+-f(x)^-$ for $f^+:=\max\{0, f\}$ and $f^-:=\max\{0,-f\}$. In the case that the RHS of $\rm (2)$ is well-defined then $$ \int_{0}^{\infty }f(x)\left(\sum_{n\geqslant 0}g_n(x)\right)\,\mathrm d x=\sum_{n\geqslant 0}\int_{0}^\infty f(x)g_n(x)\,\mathrm d x\tag3 $$ In the equality on $\rm (2)$ we just applied the monotone convergence theorem for the integral of Lebesgue. However if the RHS on $\rm (2)$ have the form $\infty-\infty$ then $\rm (1)$ is not Lebesgue integrable and is unclear if the improper integral of Riemann of $\rm (1)$ converges or if we can exchange the integral and summation sign.

---

Other way to approach the change of integral sign and summation sign is using the dominated convergence theorem. By example if $$ \sum_{n\geqslant 0}\int_{0}^\infty g_n(x)\,\mathrm d x\tag4 $$ converges, then it also does the original integral and you can exchange the summation and the integral sign.

Answer 2

$$ \int_0^\infty x^N e^{-x} \sum_{k=0}^\infty \frac{x^k}{\Gamma(N + k + 1)} f(x) \, dx \phantom{{} < +\infty} \tag 1 $$ Fubini's theorem entails that you can change the order if $$ \int_0^\infty x^N e^{-x} \sum_{k=0}^\infty \frac{x^k}{\Gamma(N + k + 1)} |f(x)| \, dx < +\infty \tag 2 $$ where $|f(x)|,$ with an absolute-value sign, appears in $(2)$ where $f(x),$ with no absolute-value sign, appears in $(1),$ and no absolute value sign need be placed elsewhere in this expression because everything else is already nonnegative.

(Fubini's theorem doesn't care about any distinction between "integrals" and "sums", the latter being integrals with respect to counting measure.)