I would like to know whether I can interchange the integral and the infinite sum as follows: $$\int_{0}^{\infty}\sum_{k = 1}^{\infty} f_k(x)\mathrm{d}x = \sum_{k = 1}^{\infty}\int_{0}^{\infty} f_k(x)\mathrm{d}x,$$ where $$f_k(x) = (-1)^{k+1}\binom{n}{k}\frac{c(ab)^{k}}{2(1+bx)^{k}}x^{k-c-1},$$ with $n \in \mathbb{C}$, $a \in (0,1]$, $b > 0$, and $c \in [0, 1]$.

Now according to Nate Eldredge's answer here, for general $f_k$ , if $\int \sum |f_k| < \infty$ or $\sum \int |f_k| < \infty$ (by Tonelli the two conditions are equivalent), then $\int \sum f_k = \sum \int f_k$.

My attempt: I tried to prove the second condition as follows: \begin{align}\sum_{k = 1}^{\infty}\int_{0}^{\infty} |f_k(x)|\mathrm{d}x &= \sum\limits_{k = 1}^{\infty} {\binom{n}{k}} \frac{c(ab)^{k}}{2} \int\limits_{0}^{\infty}\frac{x^{k-c-1}}{\left(1+b x \right)^{k}}\mathrm{d}x \\ &=\sum\limits_{k = 1}^{\infty} {\binom{n}{k}} \frac{c(ab)^{k}}{2} b^{c-k}\left[\frac{\pi}{\sin(\pi c)}\frac{\Gamma(k-c)}{\Gamma(k)\Gamma(1-c)}\right] \\ & = \frac{b^{c}}{2}\frac{\pi c}{\sin(\pi c)}\sum\limits_{k = 1}^{\infty} {\binom{n}{k}} a^{k} \left[\frac{\Gamma(k-c)}{\Gamma(k)\Gamma(1-c)}\right].\quad (1) \end{align}

I could not prove that the last expression in $(1)$ is $< \infty$.

Is my attempt on the correct track? If yes, how to prove that the RHS of $(1)$ is $< \infty$? If no, is there an another way to prove that the integral and the infinite summation can be interchanged?

  • $\begingroup$ Why don't you try the $k$ summation first? $\endgroup$ – kimchi lover Aug 15 '17 at 22:28

In order to prove that $(1)$ is finite you just have to compute the radius of convergence of a power series, namely $$ an\,\phantom{}_2 F_1\left(1-c,1-n; 2;-a\right). $$ Since by the ratio test such radius of convergence equals one, the exchange of $\int$ and $\sum$ is allowed for any $a\in(0,1)$ and the $a=1$ case can be studied as a separate instance.

The exchange of $\int$ and $\sum$ in the $a\in(0,1)$ case then leads to the identity

$$ \int_{0}^{+\infty}cx^{-1-c}\left(1-\left(\frac{1+(1-a)bx}{1+bx}\right)^n\right)\,dx=cb^c\sum_{k\geq 1}(-1)^{k+1}a^k\binom{n}{k} B(c,k-c) $$ wher $B$ is Euler's Beta function. Both sides equals $$ \frac{\pi nac b^c}{\sin(\pi c)}\phantom{}_2 F_1\left(1-c,1-n;2;a\right)=\frac{\pi n c b^c a(1-a)^{c+n}}{\sin(\pi c)}\phantom{}_2 F_1\left(c+1,n+1;2;a\right)$$ where the last identity is a consequence of Euler's transformations.

  • $\begingroup$ Thanks. Can you please elaborate on how did you reach to the Gauss hypergeometric function? Also, I notice that $a$ in the last expression (as an argument of the Gauss hypergeometric function), while in the first expression of your answer $-a$ is there as an argument of the Gauss hypergeometric function. $\endgroup$ – user389066 Aug 15 '17 at 23:29
  • $\begingroup$ @User31443: I exploited the definition of $\phantom{}_2 F_1$ through Pochhammer symbols, which are ratios of values for the $\Gamma$ function. The exchange between $a$ and $-a$ is simply due to the fact that considering $\left|f_k(x)\right|$ in place of $f_k(x)$ turns a $(-1)^{k+1}$ into a $1$, but the hypergeometric structure of the involved series stays the same. $\endgroup$ – Jack D'Aurizio Aug 15 '17 at 23:32
  • $\begingroup$ Oh, I see now! Yes, showing the convergence of radius is $1$ works. $\endgroup$ – user389066 Aug 15 '17 at 23:35
  • $\begingroup$ I thought I got the relation with the hypergeometric function, but I didn't! $\sum\limits_{k = 1}^{\infty} {\binom{n}{k}} a^{k} \left[\frac{\Gamma(k-c)}{\Gamma(k)\Gamma(1-c)}\right] = \sum\limits_{k = 1}^{\infty} {\binom{n}{k}} \binom{k-c-1}{k-1} a^{k}$. Now the right hand of the previous equation is going to $\infty$ (I checked using MATLAB). Am I missing something here? $\endgroup$ – user389066 Aug 16 '17 at 0:11
  • $\begingroup$ @User31443: do not write everything in terms of binomial coefficients, write the binomial coefficient as $\frac{\Gamma(n+1)}{k!\Gamma(n-k)}$, instead. The pattern of a $\phantom{}_2 F_1$ will become evident. And recall that $B(c,k-c)=\frac{\Gamma(c)\Gamma(k-c)}{\Gamma(k)}$, while you wrote something different. $\endgroup$ – Jack D'Aurizio Aug 16 '17 at 0:14

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