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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!

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  • $\begingroup$ 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
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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

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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.

Here:

The solution involves the hypergeometric function.

Solution

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

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  • $\begingroup$ 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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