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I wondered about this question when I've considered similar integrals with different integrand as an analogous of a formula for harmonic number.

As example that I know that can be calculated using Wolfram Alpha (it calculate the closed-form for the definite integral, and also the indefinite integral), I would like to know how get the result.

Question. Can you calculate, justifying details or hints, in closed-form $$\int_0^{\frac{\pi}{2}}\frac{1-\sqrt[18]{\cos u}}{1-\cos u}du\,?$$ Thanks in advance.

I've choosen the root $18$ since I don't know if a more simple example was in the literature.

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  • $\begingroup$ the result should be $$\sqrt{\pi } \left(\frac{\Gamma \left(\frac{37}{36}\right)}{\Gamma \left(\frac{19}{36}\right)}+\frac{\Gamma \left(\frac{55}{36}\right)}{19 \Gamma \left(\frac{37}{36}\right)}\right)-1$$ $\endgroup$ – Dr. Sonnhard Graubner Dec 30 '17 at 17:22
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    $\begingroup$ @Dr.SonnhardGraubner: the OP is probably already aware of that since he mentioned WA. $\endgroup$ – Jack D'Aurizio Dec 30 '17 at 17:37
  • $\begingroup$ I know it, but many thanks @Dr.SonnhardGraubner $\endgroup$ – user243301 Dec 31 '17 at 8:34
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This answer is similar to Jack D'Aurizio's answer, but I thought a bit more care might be needed near $u=1$. $$ \begin{align} \int_0^{\frac\pi2}\frac{1-\sqrt[n]{\cos(u)}}{1-\cos(u)}\,\mathrm{d}u &=\int_0^1\frac{1-\sqrt[n]{u}}{1-u}\frac{\mathrm{d}u}{\sqrt{1-u^2}}\tag1\\ &=\int_0^1\frac{\left(1-\sqrt[n]{u}\right)(1+u)}{\sqrt{1-u^2}^3}\,\mathrm{d}u\tag2\\ &=\frac12\int_0^1\frac{\left(1-\sqrt[2n]{u}\right)\left(1+\frac1{\sqrt{u}}\right)}{\sqrt{1-u}^3}\,\mathrm{d}u\tag3\\ &=\frac12\int_0^1\left(1-u^{\frac1{2n}}\right)(1-u)^{-\frac32}\,\mathrm{d}u\\ &+\frac12\int_0^1\left(u^{-\frac12}-u^{-\frac{n-1}{2n}}\right)(1-u)^{-\frac32}\,\mathrm{d}u\tag4\\ &=\frac12\frac{\Gamma(1)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac12\right)}-\frac12\frac{\Gamma\left(1+\frac1{2n}\right)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac12+\frac1{2n}\right)}\\ &+\frac12\frac{\Gamma\left(\frac12\right)\Gamma\left(-\frac12\right)}{\Gamma(0)}-\frac12\frac{\Gamma\left(\frac12+\frac1{2n}\right)\Gamma\left(-\frac12\right)}{\Gamma\left(\frac1{2n}\right)}\tag5\\ &=-1+\sqrt\pi\left(\frac{\Gamma\left(\frac{2n+1}{2n}\right)}{\Gamma\left(\frac{n+1}{2n}\right)}+\frac{\Gamma\left(\frac{n+1}{2n}\right)}{\Gamma\left(\frac1{2n}\right)}\right)\tag6\\ &=-1+\sqrt\pi\left(\frac{\Gamma\left(\frac{1}{2n}\right)}{2n\,\Gamma\left(\frac{n+1}{2n}\right)}+\frac{\Gamma\left(\frac{n+1}{2n}\right)}{\Gamma\left(\frac1{2n}\right)}\right)\tag7 \end{align} $$ Explanation:
$(1)$: substitute $u\mapsto\cos^{-1}(u)$
$(2)$: multiply numerator and denominator by $1+u$
$(3)$: substitute $u\mapsto\sqrt{u}$
$(4)$: distribute $1$ and $\frac1{\sqrt{u}}$ over two integrals
$(5)$: apply $(8)$
$(6)$: collect and simplify using $\Gamma\!\left(\frac12\right)=\sqrt\pi$ and $\Gamma\!\left(-\frac12\right)=-2\sqrt\pi$
$(7)$: $\Gamma(1+x)=x\,\Gamma(x)$


Justification for the non-standard Beta function limits

Let $a,b\gt0$ and $c\gt-1$. Then we can write $1=t+(1-t)$ and integrate by parts to get $$ \begin{align} &\int_0^1\left(t^{a-1}-t^{b-1}\right)(1-t)^{c-1}\,\mathrm{d}t\\ &=\int_0^1\left(t^a-t^b\right)(1-t)^{c-1}\,\mathrm{d}t +\int_0^1\left(t^{a-1}-t^{b-1}\right)(1-t)^c\,\mathrm{d}t\\ &=\frac1c\int_0^1\left((a+c)t^{a-1}-(b+c)t^{b-1}\right)(1-t)^c\,\mathrm{d}t\\[3pt] &=\frac{\Gamma(a)\,\Gamma(c)}{\Gamma(a+c)}-\frac{\Gamma(b)\,\Gamma(c)}{\Gamma(b+c)}\tag8 \end{align} $$ which can also be gotten by analytic continuation from the formula for $c\gt0$.

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  • $\begingroup$ Many thanks I am going to study your answer. $\endgroup$ – user243301 Dec 31 '17 at 8:37
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$$\int_{0}^{\pi/2}\frac{1-\sqrt[18]{\cos x}}{1-\cos x}\,dx =\int_{0}^{\pi/4}\frac{1-\sqrt[18]{2\cos^2 x-1}}{\sin^2 x}\,dx=\int_{0}^{1}\frac{1-\sqrt[18]{\frac{1-t^2}{1+t^2}}}{t^2}\,dt $$ equals $$ \frac{1}{2}\int_{0}^{1}\frac{1-\sqrt[18]{\frac{1-t}{1+t}}}{t\sqrt{t}}\,dt=\int_{0}^{1}\frac{1-\sqrt[18]{u}}{(1-u)\sqrt{1-u^2}}\,du$$ which can be managed through integration by parts and Euler's Beta function. The final outcome is $$ -1+\sqrt{\pi}\left[\frac{\Gamma\left(\frac{1}{36}\right)}{36\,\Gamma\left(\frac{19}{36}\right)}+\frac{\Gamma\left(\frac{19}{36}\right)}{\Gamma\left(\frac{1}{36}\right)}\right]\approx \frac{499}{4088}.$$ The approach is the same if $18$ is replaced by any other $m>0$: you'll get an expression involving $\Gamma\left(\frac{1}{2m}\right)$ and $\Gamma\left(\frac{1}{2}+\frac{1}{2m}\right)$.

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  • $\begingroup$ The integrals with $t$, $u$, and $w$ evaluate to $0.065457541416107517193$, However, the formula with the Gamma functions evaluates to the correct answer of $0.12206460520249458749$ $\endgroup$ – robjohn Dec 31 '17 at 0:49
  • $\begingroup$ @robjohn: thanks, now fixed. $\endgroup$ – Jack D'Aurizio Dec 31 '17 at 1:45
  • $\begingroup$ Many thanks I belive that I am going to accept the other answer, but yours is very very good. $\endgroup$ – user243301 Dec 31 '17 at 8:36

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