How can we integrate this? This question was asked in an exam but it only asked us to compare this expression's value to $\log_e2$ and to $\frac{\pi}{4}$(which I was able to solve) . I was wondering about its exact solution, or even if it's possible to calculate this exactly? $$\int_0^1\frac{dx}{1+x^\frac{\pi}{2}}$$ I just finished high school, and am at terms with basic calculus they teach at school level. If this is not solveable with basic calculus, kindly reply likewise.

  • $\begingroup$ You can get a closed form answer for this integral as an infinite series. It is probably a hypergeometric series. $\endgroup$ Aug 7, 2022 at 14:53
  • $\begingroup$ If you instead integrated from $0\to\infty$, then we could evaluate this integral exactly, without too much difficulty. $$\int_0^\infty\frac{1}{1+x^{\pi/2}}\,\mathrm{d}x=2\csc2$$ $\endgroup$
    – FShrike
    Aug 7, 2022 at 15:05
  • $\begingroup$ @CameronWilliams: I wouldn't call an infinite series a 'closed form'. $\endgroup$
    – TonyK
    Aug 7, 2022 at 15:20
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    $\begingroup$ @TonyK shrug it's an exact, explicit expression rather than an implicit expression. Closed form is not well-defined. You can call it a hypergeometric function if you want to suppress the series, but this isn't much different from doing the same for $\sin(0.1)$, for instance. $\endgroup$ Aug 7, 2022 at 15:25
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    $\begingroup$ If you do undergraduate research, there's a decent chance you'll run into hypergeometric functions. They're all over the place (for good reason). $\endgroup$ Aug 7, 2022 at 15:47

1 Answer 1


Sooner or later, you will learn that, if $\Re (a)>0$, $$\int_0^1 \frac {dx}{1+x^a}=\frac 1{2a}\left(\psi \left(\frac{a+1}{2 a}\right)-\psi \left(\frac{1}{2 a}\right)\right)$$ where appears the digamma function.

So, for $a=\frac \pi 2$

$$\int_0^1 \frac {dx}{1+x^{\frac \pi 2}}=\frac 1\pi \left(\psi \left(\frac{\pi+1}{2\pi}\right)-\psi \left(\frac{1}{\pi }\right)\right)$$ which, numerically, is $$0.75330621922217470937470011636199646707862644239580\cdots$$

  • $\begingroup$ Which representation for the digamma function do you use to obtain this? Is it a differentiation of the beta function $\endgroup$
    – FShrike
    Aug 7, 2022 at 15:14
  • $\begingroup$ @FShrike; The Gaussian hypergeometric function $\endgroup$ Aug 7, 2022 at 15:21
  • $\begingroup$ Thanks a lot, sir. @ClaudeLeibovici $\endgroup$ Aug 7, 2022 at 15:36
  • $\begingroup$ @GurmukhSingh. You are very welcome. Cheers :-) $\endgroup$ Aug 7, 2022 at 15:37

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