2
$\begingroup$

Let $0<\alpha\leq\frac{1}{2}$ a fixed real number. I wondered if it is possible to evaluate the sequence of definite integrals $$\int_0^1\left(\sum_{k=0}^n (f(x))^k\right)^{\alpha}dx\tag{1}$$ for some example of a continuous function and positive $f(x)>0$ for all $0<x<1$, and $f(x)\neq\text{constant}$.

Question. Do you know an example of evaluation of $(1)$, the evaluation $\forall n\geq 1$, for a choice of $0<\alpha\leq\frac{1}{2}$ and a function $f(x)$ satisfying previous requirements? Many thanks.

If the example that I evoke is in the literature, please refer it and I try to search and read the example from the literature.

Improve this question
$\endgroup$
3
  • $\begingroup$ Please explain what "evaluate" means. $\endgroup$
    – Max Alekseyev
    Commented Sep 14, 2019 at 13:38
  • $\begingroup$ I think that if it is possible to get an expression for each $n\geq 1$ in terms of well-known constants and/or in terms of particular values of elementary/special functions. The real $\alpha$ is a choice in your example. Many thanks for your attention @MaxAlekseyev $\endgroup$
    – user142929
    Commented Sep 14, 2019 at 13:58
  • $\begingroup$ All, if the only option is to use the definition of Riemann integral (or other) to get such definite integral, that is you cann't show an explicit example of evaluation of the sequence of previous definite integrals, explain if it is possible that one can to find such example that I evoke, or if it is impossible to get those definite integrals for any example. $\endgroup$
    – user142929
    Commented Sep 14, 2019 at 14:08

1 Answer 1

Reset to default
2
$\begingroup$

If you allow for special functions, you of course allow for quite some beasts. Generally, your expression is $$ I = \int_{0}^{1} dx \left( \sum_{k=0}^{n} (f(x))^k \right)^{\alpha } = \int_{0}^{1} dx \left( \frac{1-(f(x))^{n+1} }{1-f(x)} \right)^{\alpha } $$ Let's choose $f(x)=x^2 $ and $\alpha =1/2$, and substitute integration variable $x=\sin t$. This yields $$ I = \int_{0}^{\pi /2} dt \sqrt{1-\sin^{2n+2} t} \\ =\frac{\pi }{2}\cdot {}_{n+2}F_{n+1} \left( -\frac{1}{2},\frac{1}{2n+2},\frac{3}{2n+2},\ldots,\frac{2n+1}{2n+2};\frac{1}{n+1},\frac{2}{n+1},\ldots,\frac{n}{n+1},1;1 \right) $$

Improve this answer
$\endgroup$
2
  • $\begingroup$ Choosing $f(x)=x$ and substituting $x=\sin^2 t$ yields something similar (roughly speaking, swap the first and second sequences in the argument of the hypergeometric function, while shifting the odd sequence in the numerator up by 2). $\endgroup$
    – Michael Engelhardt
    Commented Sep 15, 2019 at 4:30
  • $\begingroup$ Many thanks for your answer. $\endgroup$
    – user142929
    Commented Sep 15, 2019 at 6:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .