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Solve: $$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx$$

I tried

$$\int_0^\frac{\pi}{2}\frac{\sqrt{\sin{x}\cos{x}}}{\cos{x}+1}dx=\sum_{k=0}^\infty(-1)^k\int_0^\frac{\pi}{2}(\sin{x})^\frac{1}{2}(\cos{x})^{k+\frac{1}{2}}=\sum_{k=0}^\infty(-1)^kB\left(\frac{3}{4},\frac{k}{2}+\frac{3}{4}\right)$$ but it did not work, did it?

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  • $\begingroup$ Wolfram says the answer is $\pi\left(\frac{\sqrt{2}}2-\frac{4\sqrt{\pi}}{\Gamma^2\left(\frac14\right)}\right)$ $\endgroup$ – Peter Foreman Apr 10 at 20:06
  • $\begingroup$ Can you post a solution, please? I would help me a lot $\endgroup$ – Reynan Henry Apr 11 at 17:21
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We are given:

$$ \int_0^{\pi/2} \frac{\sqrt{\sin x\cos x}\,dx}{1+\cos x} $$

Put $t=\tan\tfrac{x}{2}$; then

$$ \sin x = \frac{2t}{1+t^2} \quad\quad \cos x = \frac{1-t^2}{1+t^2} \quad\quad dx = \frac{2\,dt}{1+t^2} $$

Then

$$ \sqrt{\sin x\cos x} = \frac{2t(1-t^2)}{1+t^2} $$

and

\begin{align} 1 + \cos x &= 1 + \frac{1-t^2}{1+t^2} \\ &= \frac{1+t^2+1-t^2}{1+t^2} = \frac{2}{1+t^2} \end{align}

so that

$$ \int_0^{\pi/2}\frac{\sqrt{\sin x\cos x}\,dx}{1+\cos x} = \int_0^1\frac{\sqrt{2t(1-t^2)}\,dt}{1+t^2} $$

This is an elliptic integral. Here's my attempt at expressing it in terms of the Legendre canonical forms: put $t^2=v$; then $dt=2v\,dv$ and the integral becomes

\begin{align} \int_0^1 \frac{\sqrt{2t(1-t^2)}\,dt}{1+t^2} &= \int_0^1 \frac{\sqrt{2v^2(1-v^4)}(2v\,dv)}{1+v^4} \\ &= 2\sqrt2 \int_0^1 \frac{v^2(1-v^4)}{1+v^4}\frac{dv}{\sqrt{1-v^4}} \\ &= 2\sqrt2 \int_0^1 \left(\frac{i}{1+iv^2}-\frac{i}{1-iv^2}-v^2\right)\frac{dv}{\sqrt{1-v^4}} \\ &= 2\sqrt2(i\Pi(-i,i)-i\Pi(i,i)-E(i)+K(i)) \end{align}

where $K,E,\Pi$ are the first, second and third complete elliptic integrals, in terms of the modulus $k$.


Wolfram Alpha gives an exact value

$$ 2\sqrt 2(K(i)-E(i)) = -\frac{4\pi^{3/2}}{\Gamma(\tfrac{1}{4})^2} $$

which, given Peter Foreman's comment, looks right. But W|A does not give an exact value for the term involving the $\Pi$'s, so I think I made a mistake there. I'll look into that later. I'll also try and see if I can prove these values for the integrals.


Update: In

$$ E(i) - K(i) = \int_0^1\frac{v^2\,dv}{\sqrt{1-v^4}} $$

put $r=v^4$; then $dv=v^{-3}\,dr/4$ and the integral becomes

\begin{align} \int_0^1 \frac{v^2\,dv}{\sqrt{1-v^4}} &= \frac{1}{4}\int_0^1 (1-r)^{-1/2}r^{1/2}r^{-3/4}\,dr \\ &= \frac{1}{4}{\rm B}(\tfrac{1}{2},\tfrac{3}{4}) = \frac{1}{4}\frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{3}{4})}{\Gamma(\tfrac{5}{4})} \\ &= \frac{\Gamma(\tfrac{1}{2})\Gamma(\tfrac{3}{4})}{\Gamma(\tfrac{1}{4})} \end{align}

By the reflection formula,

$$ \Gamma(\tfrac{1}{4})\Gamma(\tfrac{3}{4}) = \frac{\pi}{\sin\tfrac{\pi}{4}} = \pi\sqrt 2 $$

Using next $\Gamma(\tfrac{1}{2})=\sqrt\pi$ we put everything together yielding

$$ E(i) - K(i) = \frac{\pi^{3/2}\sqrt2}{\Gamma(\tfrac{1}{4})^2} $$

I'll look into the case of the $\Pi$'s next.

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