Solving what Mathematica could not

Let $\mathcal{I}$ be the integral at hand and $B = \frac{1+A}{\sqrt{A}}\omega$. Introduce variables $y, t, \theta, z$ such that

$$y = \sin\frac{x}{2},\quad t = \tanh\theta = \frac{\cos\frac{x}{2}}{\sqrt{1+A\sin^2\frac{x}{2}}}\quad\text{ and }\quad z = e^\theta$$

Notice $$t = \sqrt{\frac{1-y^2}{1+Ay^2}} \implies y = \sqrt{\frac{1-t^2}{1+At^2}} \implies \frac{dy}{\sqrt{1+Ay^2}} = - \frac{t\sqrt{A+1}dt}{\sqrt{1-t^2}(1+At^2)} $$ and $dt = (1-t^2)d\theta$, we have $$\begin{align} \mathcal{I} &= 2\int_0^1 \sin(B\theta)\frac{dy}{\sqrt{1+Ay^2}} = 2\sqrt{A+1}\int_0^1\sin(B\theta)\frac{t\sqrt{1-t^2}}{1+At^2}\frac{dt}{1-t^2}\\ &= 2\sqrt{A+1}\int_0^\infty \frac{\sin(B\theta)\sinh\theta}{\cosh^2\theta + A\sinh^2\theta} d\theta = -i\sqrt{A+1}\int_{-\infty}^\infty \frac{e^{iB\theta}\sinh\theta}{\cosh^2\theta + A\sin^2\theta} d\theta\\ &= -i2\sqrt{A+1}\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+1)^2 + A(z^2-1)^2} dz \end{align} $$ Let $\phi = \tan^{-1}\sqrt{A}$, this can be simplified as $$ \mathcal{I} = -\frac{2i}{\sqrt{A+1}}\int_0^\infty \frac{z^{iB}(z^2-1)}{z^4 + 2\left(\frac{1-A}{1+A}\right) z^2 + 1} dz = -2i\cos\phi\int_0^\infty \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$ Consider following contour integral $$\mathcal{J(\epsilon,R)} \stackrel{def}{=} \oint_{C(\epsilon,R)} \frac{(-z)^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz \tag{*1}$$ where

  • $\arg(-z) = 0$ on negative real axis.
  • $(-z)^{iB}$ has a branch cut along positive real axis.

  • $C(\epsilon,R)$ is the contour consists of

    • $C_1$ : line segment from $\epsilon \to R$ above the positive real axis.
    • $C_2$ : circular arc $Re^{iu}$ for $u$ from $0 \to 2\pi$.
    • $C_3$ : line segment $R \to \epsilon$ below the positive real axis.
    • $C_4$ : circular arc $\epsilon e^{iu}$ for $u$ from $2\pi \to 0$.

It is easy to see in $\mathcal{J}(\epsilon,R)$,

  • the contribution from $C_1$ and $C_3$ adds up to $$(e^{\pi B} - e^{-\pi B})\int_\epsilon^R \frac{z^{iB}(z^2-1)}{(z^2+e^{2i\phi})(z^2+e^{-2i\phi})} dz$$
  • the contribution from $C_4$ vanishes as $\epsilon \to 0$.
  • Since $B > 0$ is a real number, $|(-z)^{iB}|$ is bounded from above by $e^{\pi B}$ and the contribution from $C_2$ behaves as $O(R^{-1})$ as $R \to \infty$.

Combine these, we find

$$\mathcal{I} = -i\frac{\cos\phi}{\sinh(\pi B)}\lim_{\epsilon \to 0,R \to \infty} \mathcal{J}(\epsilon,R)$$

The integrand in $(*1)$ has 4 poles inside the contour: $\; e^{i(\frac{\pi}{2} \pm \phi)}\;$ and $\;e^{i(\frac{3\pi}{2} \pm \phi)}$.

  • The residues at $e^{i(\frac{\pi}{2} \pm \phi)}$ are $\displaystyle\;(e^{-(-\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \mp\frac{e^{(\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$

  • The residues at $e^{i(\frac{3\pi}{2} \pm \phi)}$ are $\displaystyle\;(e^{-(\frac{\pi}{2} \pm \phi)B})\frac{-e^{\pm 2i\phi} - 1}{-4i e^{\pm i\phi }(-e^{\pm 2i \phi} + \cos(2\phi))} = \pm\frac{e^{(-\frac{\pi}{2}\mp \phi)B}}{4\sin\phi}$

This implies

$$\begin{align} \mathcal{I} &= \left(-i \frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi i}{4\sin\phi}\right)\left[ -e^{(\frac{\pi}{2}-\phi)B} +e^{(\frac{\pi}{2}+\phi)B} +e^{(-\frac{\pi}{2}-\phi)B} -e^{(-\frac{\pi}{2}+\phi)B} \right]\\ &= \left(\frac{\cos\phi}{\sinh\pi B}\right)\left(\frac{2\pi}{4\sin\phi}\right) (e^{\frac{\pi}{2}B} - e^{-\frac{\pi}{2}B}) (e^{\phi B} - e^{-\phi B}) = \frac{\pi}{\tan\phi}\frac{\sinh(B\phi)}{\cosh\left(\frac{\pi}{2}B\right)}\\ &= \frac{\pi}{\sqrt{A}}\frac{\sinh(B\tan^{-1}\sqrt{A})}{\cosh\left(\frac{\pi}{2}B\right)} \end{align} $$ Treating $A, B$ as two independent parameters, in the limiting case $A \to 0$, we find $$\lim_{A\to 0} \mathcal{I} = \frac{\pi B}{\cosh\left(\frac{\pi}{2}B\right)}$$

This matches what first pointed out by @nospoon in the comments.


This is not a full answer.From answer of user You're In My Eye .First, let's make the following substitutions:

$$y=\sin \frac{x}{2}$$

$$B=\frac{1+A}{\sqrt{A}} \omega$$

$$2 \int_0^1 \sin \left( B~ \text{arctanh}~\sqrt{\frac{1-y^2}{1+Ay^2}} \right) \frac{dy}{\sqrt{1+Ay^2}}$$ substitutions: $$y={\frac { \sqrt{- \left( A{t}^{2}+1 \right) \left( {t}^{2}-1 \right) }}{A{t}^{2}+1}} $$ Then we obtain: $$2\, \sqrt{A+1}\int_{0}^{1}\!{\frac {\sin \left( B{\rm arctanh} \left(t \right) \right) t}{ \left( A{t}^{2}+1 \right) \sqrt{-{t}^{2}+1}}} \,{\rm d}t $$ substitutions: $$t=\tanh \left( k \right) $$ have : $$2\,\sqrt {A+1}\int_{0}^{\infty }\!{\frac {\sin \left( Bk \right) \sinh \left( k \right) }{A \left( \cosh \left( k \right) \right) ^{2}+ \left( \cosh \left( k \right) \right) ^{2}-A}}\,{\rm d}k $$ trig identity: cosh(k)^2-sinh(k)^2 = 1 and A+1=m $$2\, \sqrt{A+1}\int_{0}^{\infty }\!{\frac {\sin \left( Bk \right) \sinh \left( k \right) }{1+ \left( A+1 \right) \left( \sinh \left( k \right) \right) ^{2}}}\,{\rm d}k \tag{1} $$ I have a simple form of integral: $$2\, \sqrt{m}\int_{0}^{\infty }\!{\frac {\sin \left( B k \right) \sinh \left( k \right) }{1+m \left( \sinh \left( k \right) \right) ^{2}}} \,{\rm d}k $$ Substitutions back $$B=\frac{1+A}{\sqrt{A}} \omega$$ to equation 1. I have: $$2\, \sqrt{A+1}\int_{0}^{\infty }\!{\frac {\sin \left(\frac{1+A}{\sqrt{A}} \omega k \right) \sinh \left( k \right) }{1+ \left( A+1 \right) \left( \sinh \left( k \right) \right) ^{2}}}\,{\rm d}k $$

Mathematica can find solution for this integral.

 A = 1/4;
 omega = 1;
 int = Normal[2*Sqrt[A + 1]*Integrate[(Sin[(1 + A)/Sqrt[A]*omega*k]*Sinh[k])/(
 1 + (A + 1)*Sinh[k]^2), {k, 0, Infinity}]]

 (*(1/1769)2 Sqrt[5] ((305 - 
  122 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 1 - (5 I)/2, 
    2 - (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
     20) Hypergeometric2F1[1, 1 - (5 I)/2, 
    2 - (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
     20) Hypergeometric2F1[1, 1 - (5 I)/2, 2 - (5 I)/2, (1 - 2 I)/
    Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[1, 1 - (5 I)/2, 
    2 - (5 I)/2, (1 + 2 I)/Sqrt[5]]) + (5 + 
  2 I) (61 ((1/5 + (3 I)/20) Hypergeometric2F1[1, 1 + (5 I)/2, 
       2 + (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
        20) Hypergeometric2F1[1, 1 + (5 I)/2, 
       2 + (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
        20) Hypergeometric2F1[1, 1 + (5 I)/2, 2 + (5 I)/2, (
       1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[1, 
       1 + (5 I)/2, 2 + (5 I)/2, (1 + 2 I)/Sqrt[5]]) + (2 + 
     5 I) ((6 + 
        5 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 3 - (5 I)/2, 
          4 - (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
           20) Hypergeometric2F1[1, 3 - (5 I)/2, 
          4 - (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
           20) Hypergeometric2F1[1, 3 - (5 I)/2, 4 - (5 I)/2, (
          1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[
          1, 3 - (5 I)/2, 4 - (5 I)/2, (1 + 2 I)/Sqrt[5]]) - (6 - 
        5 I) ((1/5 + (3 I)/20) Hypergeometric2F1[1, 3 + (5 I)/2, 
          4 + (5 I)/2, -((1 + 2 I)/Sqrt[5])] + (1/5 - (3 I)/
           20) Hypergeometric2F1[1, 3 + (5 I)/2, 
          4 + (5 I)/2, -((1 - 2 I)/Sqrt[5])] + (1/5 - (3 I)/
           20) Hypergeometric2F1[1, 3 + (5 I)/2, 4 + (5 I)/2, (
          1 - 2 I)/Sqrt[5]] + (1/5 + (3 I)/20) Hypergeometric2F1[
          1, 3 + (5 I)/2, 4 + (5 I)/2, (1 + 2 I)/Sqrt[5]]))))*)

Here is a sketch for a residue-free solution:

First, consider the formula $$\int_0^{\infty} \frac{\cos( a x)}{\cosh(\frac{\pi}{2} x)}dx = \operatorname{sech} a. \tag{1}$$

Since $\displaystyle \,\,\sin(a)\sin(b) = \frac12 \cos(a-b)-\frac12 \cos(a+b), \,\,$ we obtain $$ \int_0^{\infty} \frac{\sin( a x) \sin(b x)}{\cosh(\frac{\pi}{2} x)}dx = \frac12 \operatorname{sech}(a-b) - \frac12 \operatorname{sech}(a+b). \tag{2}$$

Now, using a fourier inversion argument $\displaystyle\left( f(a)=\int_0^{\infty} g(x)\sin(a x) dx \iff \frac{\pi}{2} g(a) = \int_0^{\infty} f(x) \sin(a x) dx \right),\,$ we obtain $$\int_0^{\infty} \sin( b x) ( \operatorname{sech}(a-x) - \operatorname{sech}(a+x)) dx = \pi \cfrac{ \sin(a b)}{\cosh( \frac{\pi}{2} b)}. \tag{3}$$

Finally, letting $a \mapsto i \tan^{-1} \sqrt{a},\,$ and noting that $$\operatorname{sech}(a-b) - \operatorname{sech}(a+b) = 2 \frac{\sinh a}{\cosh^2 a} \,\frac{\sinh b}{\cosh^2 b}\, \frac1{1-\tanh^2 a \, \tanh^2 b},$$

we find that

$$\frac{\pi}{\sqrt{a}} \cfrac{\sinh( b \tan^{-1} \sqrt{a} )}{\cosh( \frac{\pi}{2} b)} = 2 \sqrt{1+a} \int_0^{\infty} \cfrac{\sin(b x) \sinh x}{\cosh^2 x + a \, \sinh^2 x} dx, \tag{4}$$

which is exactly your integral, with $b= \frac{1+a}{\sqrt{a}} \omega,$ and the substitution $$\tanh^{-1} \frac{\cos \frac{x}{2}}{\sqrt{1+ a \sin^2 \frac{x}{2}}} \mapsto x. \tag{5}$$

Tags:

Integration

Elliptic Integrals

Definite Integrals

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