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This is the sequel of my previous question

$$I(a)=\int_{0}^{\infty}\frac{\arctan (a\,\sin^2x)}{x^2}dx$$ I want to use differentiation under the integral sign with respect to parameter "a" but so far without success.

Any hint?

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    $\begingroup$ Do you mean $(\arctan a) (\sin^2 x)$ or $\arctan(a\sin^2 x)$? I'm guessing the latter, since the former is trivial, but the normal way to read this is the former. $\endgroup$ – Thomas Andrews Jul 23 '12 at 14:45
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    $\begingroup$ @ThomasAndrews Edited $\endgroup$ – Martin Gales Jul 23 '12 at 14:51
  • $\begingroup$ interestingly this can be also tackled by residue methods. if u are interested i write something $\endgroup$ – tired Oct 25 '16 at 23:34
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Thanks for the nice question.

The answer is $$ I(a) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}} $$ The sketch of the proof: expand $\arctan$ in series, and integrate term-wise (can do this for small enough $a$, since the sine is bounded): $$ \arctan\left(a \sin^2(x)\right) = \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \sin^{4n+2}(x) $$ This gives $$ \int_0^\infty \frac{\sin^{4n+2}(x)}{x^2} \mathrm{d} x = \frac{1}{\binom{2n}{\tfrac{1}{2}}} = \frac{\sqrt{\pi}}{2} \cdot \frac{\Gamma(2n+\frac{1}{2})}{(2n)!} $$ The summation is easy, since the summand is a hypergeometric term: $$ I(a) = \frac{\sqrt{\pi}}{2} \sum_{n=0}^\infty \frac{(-1)^n a^{2n+1}}{2n+1} \frac{\Gamma(2n+\frac{1}{2})}{(2n)!} = \frac{\pi a}{2} \cdot {}_2F_1\left(\frac{1}{4}, \frac{3}{4}; \frac{3}{2}; -a^2\right) = \frac{\pi}{\sqrt{2}} \cdot \frac{a}{ \sqrt{1 + \sqrt{1+a^2}}} $$


Added: The hard part is to prove that $S_n = \int_0^\infty \frac{\sin^{4n+2}(x)}{x^2} \mathrm{d} x$ is a hypergeometric term as claimed above. This can be done using: $$\begin{eqnarray} \sin^{4n+2}(x) &=& \left(\frac{\mathrm{e}^{ix} - \mathrm{e}^{-i x}}{2i}\right)^{4n+2} = -\frac{1}{4} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \mathrm{e}^{i (4n+2-2m)x} \\ &\stackrel{\text{symmetry}}{=}& -\frac{1}{4} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \underbrace{\cos((4n+2-2m)x)}_{1-2 \sin^2((2n+1-m)x)} \\ &=& \frac{1}{2} \cdot \frac{1}{16^n} \sum_{m=0}^{4n+2} \binom{4n+2}{m} (-1)^m \sin^2((2n+1-m)x) \\ &\stackrel{\text{symmetry}}{=}& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \sin^2((2n+1-m)x) \end{eqnarray} $$ Now: $$\begin{eqnarray} S_n &=& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \int_0^\infty \frac{\sin^2\left((2n+1-m) x\right)}{x^2} \mathrm{d} x \\ &=& \frac{1}{16^n} \sum_{m=0}^{2n} \binom{4n+2}{m} (-1)^m \frac{\pi}{2} \left(2n+1-m\right) \\ & \stackrel{m \to 2n-m}{=}& \frac{1}{16^n} \frac{\pi}{2} \sum_{m=0}^{2n} \binom{4n+2}{2n+2+m} (-1)^m \left(m+1\right) \end{eqnarray} $$ The latter sum readily yields to telescoping method, establishing the claim: $$ S_n = \frac{\pi}{2} \cdot \frac{n+1 }{4 n+1} \cdot \frac{1}{16^n} \binom{4 n+2}{2 n+2} = \frac{\sqrt{\pi}}{2} \frac{\Gamma\left(2n+\frac{1}{2}\right)}{(2n)!} $$

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  • $\begingroup$ Did you solve this using Wolfram Alpha? $\endgroup$ – Chibueze Opata Jul 23 '12 at 16:13
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    $\begingroup$ @Sasha Very impressive! Thanks! $\endgroup$ – Martin Gales Jul 24 '12 at 14:02
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One more solution: using $\sum_{n=-\infty}^{\infty}\left(x+2\pi n\right)^{-2}=\frac{1}{4}\csc^{2}\frac{x}{2}$ and$$\sum_{n\ge0}C_{n}x^{n}=\frac{1-\sqrt{1-4x}}{2x}\implies\sum_{n\ge0}C_{2n}x^{2n}=\frac{\sqrt{1+4x}-\sqrt{1-4x}}{4x}$$(by taking the even part) in terms of the Catalan numbers $C_{n}=\frac{\left(2n\right)!}{n!\left(n+1\right)!}$,$$\begin{align}I\left(a\right)&=\frac{1}{2}\int_{-\infty}^{\infty}\frac{\arctan\left(a\sin^{2}x\right)}{x^{2}}dx\\&=\frac{1}{2}\int_{0}^{2\pi}\sum_{n=-\infty}^{\infty}\frac{\arctan\left(a\sin^{2}x\right)}{\left(x+2\pi n\right)^{2}}dx\\&=\frac{1}{8}\int_{0}^{2\pi}\arctan\left(a\sin^{2}x\right)\csc^{2}\frac{x}{2}dx\\&=\frac{1}{4}\int_{0}^{\pi/2}\arctan\left(a\sin^{2}x\right)\left(\csc^{2}\frac{x}{2}+\sec^{2}\frac{x}{2}\right)dx\\&=\int_{0}^{\pi/2}\arctan\left(a\sin^{2}x\right)\csc^{2}xdx\\&=\sum_{n\ge0}\frac{\left(-1\right)^{n}}{2n+1}a^{2n+1}\int_{0}^{\pi/2}\sin^{4n}xdx\\&=\pi a\sum_{n\ge0}C_{2n}\left(-\frac{a^{2}}{16}\right)^{n}\\&=\frac{\pi}{i}\left(\sqrt{1+ia}-\sqrt{1-ia}\right)\\&=\frac{\pi}{\sqrt{2}}\frac{a}{\sqrt{1+\sqrt{1+a^{2}}}}.\end{align}$$

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Good evening,

I found a simpler faster way to deal with this integral :

1/ The numerator being a pair function of Sin(x) one can easily show that the integral is equal to :

F(a)=Integral from 0 to Pi/2 Arctan(a.(Sin(x)^2))) / (Sin(x)^2) dx

2/ Then we apply Feynman’trick to F(a) and get :

F’(a)=Integral from 0 to Pi/2 dx / (1+ a^2. (Sin(x))^4

3/ Then we use t = Tan(x) and get :

F’(a)=Integral ftom 0 to inf (1+t^2) dt / (t^4(1+a^2) + 2t^2 +1)

4/ Then we use Schlomilch master formula and get :

F’(a)=Pi / (2^(1/2)) . (1+(1+a^2)^(1/2)) / (1+a^2)^(1/2)

5/ Then we integrate F’(a) and get :

F(a) = Pi / (2^(1/2)) . a / (1+(1+a^2)^(1/2))^(1/2)

     = Pi / (2^(1/2) . ((1+a^2)^(1/2) - 1)^(1/2)

fjaclot

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  • $\begingroup$ Please use MathJax. $\endgroup$ – Martin Gales Nov 27 at 18:34

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