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While contending with a certain Fourier series, I stumbled on an incredibly simple evaluation (numerically) of a slightly complicated-looking sin-integral.

So, I wish ask:

Question. Is this really true? If so, any proof? $$I:=\int_0^{\frac{\pi}2}\frac{\sin x}{1+\sqrt{\sin 2x}}\,dx=\frac{\pi}2-1.$$

ADDED. I'm an experimentalist and I find many many results. Some I could find being discovered earlier after checking the literature. For others, either I don't find them easily or I might be tired of looking and hope someone else points them out to me. I'm mostly interested in sharing and having fun, not seeking recognition of any sort. However, one thing is for sure: I don't give oxygen to rude comments.

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    $\begingroup$ Further substituting $y = 2t/(1+t^2)$ yields a rational function of $t$ and $\sqrt{1+t^2}$, and then yet another substitution $t=(1-u^2)/2u$ makes it a rational function of $u$, thus reducing to a problem known to be solvable. $\endgroup$ Jun 16, 2017 at 20:30
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    $\begingroup$ @NoamD.Elkies substituting $y=2t/(1+t^2)$ I get $\sqrt{t-t^3}$ involved, not $\sqrt{1+t^2}$. I doubt, really, that indefinite integral is expressed in elementary functions. $\endgroup$ Jun 16, 2017 at 21:07
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    $\begingroup$ Yes, I wondered too: the functions $u := \sin x$ and $v := \sqrt{\sin 2x}$ satisfy $v^4 = 2u^2(1-u^2)$, which seems to define a genus $1$ curve (actually a twist of $y^2=x^3-x$, up to birational equivalence). So that seems to preclude the possibility of obtaining a rational function for the integrand. $\endgroup$
    – RP_
    Jun 16, 2017 at 21:39
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    $\begingroup$ \begin{align} \frac{1}{2} \left(\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{\tan x}}\right)-\tan ^{-1}\left(\frac{\sqrt{2}}{\sqrt{\tan x}}+1\right)\right)+\frac{1}{2\sqrt{2}}\log \left(\frac{\sqrt{\tan x}+1}{\sqrt{\tan x}-1}\right)\\+\frac12\frac{1-\sqrt{\sin 2 x}}{\cos x-\sin x}-\frac{1}{\sqrt{2}}\tanh ^{-1}\left(\frac{\tan \left(\frac{x}{2}\right)+1}{\sqrt{2}}\right)+\frac{1}{4} \log \left(\frac{\sqrt{\tan x}+\sqrt{\cot x}-\sqrt{2}}{\sqrt{\tan x}+\sqrt{\cot x}+\sqrt{2}}\right) \end{align} $\endgroup$
    – Nemo
    Jun 16, 2017 at 23:19
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    $\begingroup$ @T.Amdeberhan this is problem 11961 from Feb 2017 issue of American Mathematical Monthly. $\endgroup$
    – Nemo
    Jul 8, 2017 at 11:04

3 Answers 3

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We have

\begin{align} & 2\int_0^{\pi/2}\frac{\sin x}{1+\sqrt{\sin 2x}} \, dx=\int_0^{\pi/2}\frac{\sin x+\cos x}{1+\sqrt{\sin 2x}} \, dx=\frac12\int_0^\pi\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy \\[6pt] = {} &\int_0^{\pi/2}\frac{\sqrt{1+\sin y}}{1+\sqrt{\sin y}} \, dy =\int_0^1\frac{\sqrt{1+t}}{(1+\sqrt{t})\sqrt{1-t^2}} \, dt=\int_0^1\frac{dt}{(1+\sqrt{t})\sqrt{1-t}} \\[6pt] = {} &2\int_0^{\pi/2}\frac{\cos z}{1+\cos z} \, dz=\pi-2\int_0^{\pi/2}\frac1{1+\cos z} \,dz= \pi-2\tan\frac{z}2\bigg|_0^{\pi/2}=\pi-2, \end{align} where we used substitutions $y=2x$, $t=\sin y$, $t=\cos^2 z$.

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    $\begingroup$ That's a nice maneuvering. $\endgroup$ Jun 16, 2017 at 21:28
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    $\begingroup$ And formatting. ( ͡° ͜ʖ ͡°) $\endgroup$ Jun 18, 2017 at 12:11
  • $\begingroup$ Can I ask how you got $2 \sin x = \sin x + \cos x $? $\endgroup$ Sep 14, 2021 at 20:27
  • $\begingroup$ @NazmulHasanShipon I did not. What I did is $\int_0^{\pi/2} \sin x f(\sin 2x)dx=\int_0^{\pi/2} \sin (\pi/2-x) f(\sin 2(\pi/2-x))dx=\int_0^{\pi/2} \cos x f(\sin 2x)dx$ for $f(t)=1/(1+\sqrt{t})$ $\endgroup$ Sep 14, 2021 at 21:09
  • $\begingroup$ @FedorPetrov did you make a substitution for $x$ by $\frac{\pi}{2}−x$ in your above comment? $\endgroup$ Sep 15, 2021 at 8:49
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This is not as crisp as Fedor's solution (which I am accepting), but it might help to see alternative techniques. In fact, I very much welcome others to join the effort (for pedagogical reasons).

From the geometric series expansion, $\frac1{1+\sqrt{\sin 2x}}=\sum_{n=0}^{\infty}(-1)^n\sin^{\frac{n}2}2x$. From the Euler's beta, \begin{align} \int_0^{\frac{\pi}2}\sin x\,\sin^{\frac{n}2}2x\,dx=2^{\frac{n}2}\int_0^{\frac{\pi}2}\sin^{\frac{n}2+1}x\,\cos^{\frac{n}2}x\,dx=\binom{\frac{n}2+\frac12}{\frac{n}2}^{-1}. \end{align} Therefore, we have \begin{align} I=\sum_{n=0}^{\infty}(-1)^n\binom{\frac{n}2+\frac12}{\frac{n}2}^{-1} &=\sum_{n=0}^{\infty}\left[\binom{n+\frac12}n^{-1}-\binom{n+1}{n+\frac12}^{-1}\right] \\ &=\sum_{n=0}^{\infty}\left[\frac{2^{2n}}{(2n+1)\binom{2n}n}-\frac{\pi}2\frac{\binom{2n+2}{n+1}}{2^{2n+2}}\right]. \end{align} Letting $a_n:=2^{2n}\binom{2n}n^{-1}$, we may rewrite $I=\sum_{n\geq0}\left[\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\right]+\frac{\pi}2$. Stirling's approximation shows that $a_n\sim\sqrt{\pi n}$ and hence $\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\sim\frac1{n^{\frac32}}$. This ensures the integral $I$ exists, despite the fact that both $\sum\frac{a_n}{2n+1}$ and $\sum\frac{\pi}{2a_n}$ diverge, individually. On the other hand, we know $$ f(x):=\sum_{n\geq 0} \frac{a_nx^{2n}}{2n+1} =\frac{\sin^{-1}x} {x\sqrt{1-x^2}} \qquad \text{and} \qquad g(x):=\sum_{n\geq 0}\frac{x^{2n}}{a_n} =\frac{1}{\sqrt{1-x^2}}. $$ Invoking Abel's Theorem and using L'Hopital's Rule, compute that $$\sum_{n\geq0}\left[\frac{a_n}{2n+1}-\frac{\pi}{2a_n}\right]=\lim_{x\to 1-}\left(f(x)-\frac{\pi}2 g(x)\right)=-1.$$ In the end, we arrive at $I=\frac{\pi}2-1$ as required.

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    $\begingroup$ Assuming the problem is a pesky mosquito, I'd have to say Fedor used a mosquito swatter and you used a hydrogen bomb to take care of it. :D +1 $\endgroup$ Jun 17, 2017 at 5:22
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Here is another approach: $$\begin{align}2I&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{\sin 2t}}\,dt\\&=\int_{0}^{\pi/2}\frac{\sin t+\cos t}{1+\sqrt{1-(\sin t-\cos t)^2}}\,dt\\& =\underbrace{ \int_{-\pi/2}^{\pi/2}\frac{\cos u}{1+\cos u}\,du}_{\sin t-\cos t=\sin u }\\&=2\int_0^{\pi/2}\left(1-\dfrac{1}{1+\cos u}\right)\,du\end{align}$$ $$\boxed{I=\frac{\pi}{2}-1}$$

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