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Show that $\int_0^\frac{\pi}{2}\sqrt{\sin x} dx \times \int_0^\frac{\pi}{2}\frac{1}{\sqrt{\sin x}} dx =\pi $

My teacher gave this question to solve but I was unable to solve it.

I think there is surely any property of definite integral which I'm missing. I'm trying not to use exponential integral or any other special function.

I tried the following method: $$\int_0^\frac{\pi}{2}\sqrt{\sin x} dx \times \int_0^\frac{\pi}{2}\frac{1}{\sqrt{\sin x}} dx$$ $$\int_0^\frac{\pi}{2}\sqrt{\sin x} \times \frac{1}{\sqrt{\sin x}} dx$$ $$\int_0^\frac{\pi}{2}dx = \frac{\pi}{2} $$

This is of course not true. What should I do with this? Kindly help me.

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  • $\begingroup$ Are you allowed Beta & Gamma functions, or those too advanced? $\endgroup$
    – J.G.
    Commented Jan 13, 2022 at 15:11
  • $\begingroup$ dlmf.nist.gov/5.12 5.12.2 $\endgroup$
    – Svyatoslav
    Commented Jan 13, 2022 at 15:14
  • $\begingroup$ I am not convinced this is possible to do without using special functions. $\endgroup$
    – Angel
    Commented Jan 13, 2022 at 15:21
  • $\begingroup$ The integrals involved are elliptic so no means to solve this directly by elementary ways. We can put instead as follows: $$\int_0^\frac{\pi}{2}(\sqrt{ x}-\frac{x^{5/2}}{12}+O(x^{11/2})) dx \times \int_0^\frac{\pi}{2}\frac{1}{\sqrt x}+\frac{x^{3/2}}{12}+O(x^{7/2})) dx$$ $\endgroup$
    – Piquito
    Commented Jan 13, 2022 at 15:50
  • $\begingroup$ @Piquito may I know what's 'O' here? $\endgroup$
    – user983440
    Commented Jan 13, 2022 at 15:51

2 Answers 2

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Here's an answer that does not rely on special functions. As in Mokrane's original approach, we combine the two integrals to obtain a double integral (the same method is also used in this answer to a similar question).

We have $$ P \equiv \int \limits_0^{\pi/2} \sqrt{\sin(x)} \, \mathrm{d} x \int \limits_0^{\pi/2} \frac{\mathrm{d} y}{\sqrt{\sin(y)}} = \int \limits_0^{\pi/2} \int \limits_0^{\pi/2} \sqrt{\frac{\sin(x)}{\sin(y)}} \, \mathrm{d} x \, \mathrm{d} y \ . $$ Letting $u = \sin(x)$ and $v = \sin(y)$, we find \begin{align} P &= \int \limits_0^1 \int \limits_0^1 \frac{\sqrt{\frac{u}{v}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v = \frac{1}{2} \int \limits_0^1 \int \limits_0^1 \frac{\sqrt{\frac{u}{v}} + \sqrt{\frac{v}{u}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v \\ &= \int \limits_0^1 \int \limits_0^v \frac{\sqrt{\frac{u}{v}} + \sqrt{\frac{v}{u}}}{\sqrt{1-u^2}\sqrt{1-v^2}} \, \mathrm{d} u \, \mathrm{d} v \, , \end{align} where we exploited the symmetry of the integrand to simplify the result. Now the substitution $u = v t^2$ in the inner integral yields \begin{align} P &= 2 \int \limits_0^1 \int \limits_0^1 \frac{(1 + t^2) v}{\sqrt{1 - t^4 v^2}\sqrt{1-v^2}} \, \mathrm{d} t \, \mathrm{d} v = 2 \int \limits_0^1 (1+t^2) \int \limits_0^1 \frac{v}{\sqrt{1 - t^4 v^2}\sqrt{1-v^2}} \, \mathrm{d} v \, \mathrm{d} t \\ &\overset{(*)}{=} 2 \int \limits_0^1 (1+t^2) \frac{\operatorname{artanh}(t^2)}{t^2} \, \mathrm{d} t = 2 \int \limits_0^1 \left(1 + \frac{1}{t^2}\right) \operatorname{artanh}(t^2) \, \mathrm{d} t \overset{\text{IBP}}{=} 2 \int \limits_0^1 \left(\frac{1}{t} - t\right) \frac{2 t}{1-t^4} \, \mathrm{d} t \\ &= 4 \int \limits_0^1 \frac{\mathrm{d} t}{1+t^2} = 4 \arctan(1) = \pi \, . \end{align}


Proof of $(*)$:

The Euler substitution $w = \sqrt{\frac{1-v^2}{a^{-2}-v^2}}$ yields $$ \int \limits_0^1 \frac{v}{\sqrt{1-a^2 v^2}\sqrt{1-v^2}} \, \mathrm{d} v = \frac{1}{a} \int \limits_0^a \frac{\mathrm{d}w}{1-w^2} = \frac{\operatorname{artanh}(a)}{a} $$ for $a \in (0,1)$.

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$$ I = \int_0^{\pi/2} \sqrt{\sin x} dx \int_0^{\pi/2} \dfrac{1}{\sqrt{\sin x}} dx $$ By substituting $u = \sin x$, we get to the product of the two following simple integrals: $$ P = \int_0^1 \sqrt{\dfrac{u}{1-u^2}}du $$ and $$ Q = \int_0^1 \dfrac{1}{\sqrt{v(1-v^2)}}dv $$ Now, by putting $w = u^2$ in $P$ then $w = v^2$ in $Q$, we bring these two integrals into Euler's Beta function : $$ P = \dfrac{1}{2}B\left(\dfrac{3}{4},\dfrac{1}{2}\right) $$ $$ Q = \dfrac{1}{2}B\left(\dfrac{1}{4},\dfrac{1}{2}\right) $$ where $B(x,y) = \int_0^1 t^{x-1}(1-t)^{y-1} dt$, which in turn can be brought into Euler's Gamma function through the formula $B(x,y) = \dfrac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$. Hence, we get $$ I = \Gamma\left(\dfrac{1}{2}\right)^2 = \pi. $$

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    $\begingroup$ I would point out that there is no point in doing the combination to a double integral with this approach. You can just do $u$-substitution into each integral and get to the beta functions faster. The fact that the double integral is not used in this way makes me sense there is a different way to evaluate. $\endgroup$
    – Paul
    Commented Jan 13, 2022 at 15:55
  • $\begingroup$ Yes, definitely. $\endgroup$
    – Mokrane
    Commented Jan 13, 2022 at 17:19

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