I am trying to show that $$\int_0^1 \frac{x^{p-1}}{1+x^q} dx = \sum_{n=0}^\infty \frac{(-1)^n}{p+nq}$$ whereas $p,q >0$. I thought, I might just use the geometric series and then interchange sum and integral, so I would have:

$$\int_0^1 \frac{x^{p-1}}{1+x^q} dx = \int_0^1 x^{p-1} \sum_{n=0}^\infty (-x^q)^n$$ That way, if interchanging the sum with the integral is valid I would arrive at my goal pretty easily. However, I am not quite sure if I am allowed to do this.

I know for a fact that the geometric series is uniformly convergent for every compact subset $K \subset [-1,1]$. The integrals of the partial sums seem to exist as well. But is compact convergence enough in this case? I know that I can interchange the sum and the integral if my partial sums converge uniformly on the whole interval (in this case $[0,1]$), but this does not seem to apply here. I also thought about using dominated convergence, but the upper boundary $g(x) :=\frac{1}{1-x^q}$ would not be an integrable function on this interval...

Any help would be greatly appreciated!

  • $\begingroup$ This won’t work for $ q >1$. $\endgroup$ – Paul Nov 11 '17 at 14:09
  • $\begingroup$ You have $x^q$ instead of $xq$ right? $\endgroup$ – Shashi Nov 11 '17 at 14:16
  • $\begingroup$ @Shashi Yes, I am correcting it. thanks $\endgroup$ – Markus Peschl Nov 11 '17 at 14:16
  • $\begingroup$ And why is the mentioned $g(x) $ not integrable? It seems very integrable to me. $\endgroup$ – Shashi Nov 11 '17 at 14:18
  • $\begingroup$ The dominated convergence theorem allows the interchange (use $x^{p-1}$ as the dominating function). That is, if you use the Lebesgue integral. If you use the Riemann integral, you need to do a bit more work since the DCT doesn't fully hold for the Riemann integral. $\endgroup$ – Daniel Fischer Nov 11 '17 at 14:18

It is the correct idea, the sequence $\int_0^{1-{1\over n}}{x^{p-1}\over{1+x^q}}dx$ is an increasinig and bounded sequence, this implies that

$\int_0^{1-{1\over n}} x^{p-1} \sum_{n=0}^\infty (-x^q)^n$ is also increasing and bounded so it converges towards its sup


Here are some hints which may guide you in solving this problem. On another note, would love to know which textbook/website this problem came from - really interesting.


The solution involves the hypergeometric function.


More information on the hypergeometric function

Another way to write the series

Another way to write the function as a sum

Taylor expansion at t=0

Taylor expansion at $t=0$ enter image description here

  • $\begingroup$ Unfortunately, this problem was given to me, so if it was taken from a textbook, I could not tell you where it is from. $\endgroup$ – Markus Peschl Nov 28 '17 at 20:55

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