# Switching the Order of Summation and Integration: Incomplete Gamma Function

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!

• you can do it as far the elements that are summed are non-negative. it is an application of the monotone convergence theorem Nov 1, 2019 at 2:14
• @Masacroso Okay, so we can take the supremum needed in the monotone convergence theorem to be the value of the incomplete Gamma function? Nov 1, 2019 at 2:20

$$\int_0^\infty x^N e^{-x} \sum_{k=0}^\infty \frac{x^k}{\Gamma(N + k + 1)} f(x) \, dx \phantom{{} < +\infty} \tag 1$$ 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.)

• So then, according to Fubini's theorem, I can write, $f(x, y) = \sum_{k=0}^{\infty} \frac{y^{k}}{\Gamma(N+k+1)} \cdot x^{N} e^{-x} f(x)$. So I just need to show that both: $\sum_{k=0}^{\infty} \frac{y^{k}}{\Gamma(N+k+1)}$ converges and $\int_{0}^{\infty} x^{N} e^{-x} f(x) dx$ converges, right? So, if I know that $\int_{0}^{\infty} f(x) dx \leq M$, $M >0$ does that mean that $\int_{0}^{\infty} x^{N} e^{-x} f(x) dx$ converges as well? I have a feeling the sum converges because the Gamma function grows much faster than the power. Nov 1, 2019 at 17:55
• And when I say converge, I just mean that the sum and integral are finite, should that suffice? Nov 1, 2019 at 18:11
• @TheDude : You need an integral of an ABSOLUTE VALUE to be finite. $\qquad$ Nov 1, 2019 at 20:46
• Okay, so if $\int_{0}^{\infty} |f(x)|dx \leq M < \infty$ and $M > 0$, then $\int_{0}^{\infty} x^{N} e^{-x} f(x) dx$ converges? My intuition is because $x^{N} = o(e^{-x})$, we can say this. Nov 1, 2019 at 22:37
• @TheDude : To say that $\int_0^\infty x^N e^{-x} f(x)\, dx \text{ “converges''}$ is a statement that I am inclined to construe as meaning that $\int_0^\infty \large| x^N e^{-x} f(x)\large| \, dx < +\infty$ (where you will observe that this last expression has an absolute value sign and the former expression, said to "converge", does not). And that is correct since $\int_0^\infty x^N e^{-x}\,dx,$ (an integral of a non-negative function) is finite. $\qquad$ Nov 2, 2019 at 4:00

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.

• Okay, well $g_{n} = \frac{x^{n}}{\Gamma(N+n+1)}$, but $\int_{0}^{\infty} \frac{x^{n}}{\Gamma(N+n+1)} dx = \frac{1}{\Gamma(N+n+1)} \frac{1}{n+1} x^{n+1} \Big |_{0}^{\infty} = \infty$ So in this case, it doesn't converge, right? Nov 1, 2019 at 18:01
• But if the right most integral in the RHS of (2) is zero, for instance if $f(x) \geq 0$, does it count? Nov 1, 2019 at 18:08
• in your case $g_n(x)=\frac{x^{N + n}}{e^{x}\Gamma (N +k+1) }$. For your second question: yes, it count, that is, if $f^-=0$ then everything simplifies Nov 1, 2019 at 23:15
• Okay, well then it definitely will converge then. Thanks for your help! Nov 2, 2019 at 19:45