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I want to transform one integral representations of the Besselfunction of first kind of zeroth order into another $$ J_0(x) = \frac 2 \pi \int_1^\infty \frac{\sin(xt)}{\sqrt{t^2-1}}\,\mathrm dt = \frac 1\pi \int_0^\pi \cos\big(x\sin(\varphi)\big)\,\mathrm d\varphi $$ I want to do it with integration techniques (substitution, partial integration, ...) and not just show the equivalence via series representation of the Bessel function or the Bessel's differential equation.

I tried a substitution of $$ t = \tan\big(\frac \varphi 2\big) + 1\\ \mathrm dt = \frac{1}{2\cos\big(\frac \varphi 2\big)^2} \mathrm d\varphi\\ \varphi = 2\arctan(t-1)\\ $$ to transform the limits from $1 ... \infty$ to $0 ... \pi$. It follows $$ J_0(x) = \frac 2 \pi \int_0^\pi \frac{\sin\big[x\big(\tan\big(\frac \varphi 2\big)+1\big)\big]}{2\cos\big(\frac \varphi 2\big)^2\sqrt{\big(\tan\big(\frac \varphi 2\big) + 1\big)^2 -1}}\,\mathrm d\varphi\\ = \frac 1 \pi \int_0^\pi \frac{\sin\big[x\big(\tan\big(\frac \varphi 2\big)+1\big)\big]}{\cos\big(\frac \varphi 2\big)^2\sqrt{\tan\big(\frac \varphi 2\big)^2 + 2\tan\big(\frac \varphi 2\big)}}\,\mathrm d\varphi $$ With this, the integrals run over the same range, however the integrands are still completely different (plotting the integrands shows completely different curves but integration gives valid results according to mathematica).

How can I transform this complicated integrand into the simpler one? Or is there a better possibility for a substitution, ...?

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  • $\begingroup$ These integrals in both sides are conclusions of generating function for Bessel, it seems hard to find analytical connection from one to another! $\endgroup$ – Nosrati Nov 30 '17 at 16:06
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We can transform the first representation \begin{align} I_1&=\frac 2 \pi \int_1^\infty \frac{\sin(xt)}{\sqrt{t^2-1}}\,\mathrm dt\\ &=\frac 2 \pi \int_0^\infty \sin x\cosh u\,du \\ &=\frac 2 \pi\Im \int_0^\infty \exp(ix\cosh u)\,du \end{align} In the complex plane, introducing the rectangular contour made by a vertical segment $u=iz, 0<z<\tfrac{\pi}{2}$ and an horizontal part $u=i\pi/2+s, 0<s<\tfrac{\pi}{2}$, the residue theorem shows that \begin{align} \int_0^\infty \exp(ix\cosh u)\,du &=i\int_0^{\tfrac{\pi}{2}}\exp\left( ix\cosh iz \right)\,dz+\int_0^\infty\exp\left( ix\cosh(s+\tfrac{\pi}{2}) \right)\,ds\\ &=i\int_0^{\tfrac{\pi}{2}}\exp\left( ix\cos z \right)\,dz+\int_0^\infty\exp\left( -x\sinh(s) \right)\,ds \end{align} As the second integral of the last expression is real, \begin{align} I_1&=\frac 2 \pi \Im\left[i\int_0^{\tfrac{\pi}{2}}\exp\left( ix\cos z \right)\,dz\right]\\ &=\frac 2 \pi\int_0^{\tfrac{\pi}{2}}\cos\left( x\cos z \right)\,dz\\ &=\frac 2 \pi\int_0^{\tfrac{\pi}{2}}\cos\left( x\sin z \right)\,dz\\ &=\frac 1 \pi\int_0^{\pi}\cos\left( x\sin z \right)\,dz \end{align} In the last two expressions, we have changed $z\to \pi/2-z$ and have used the symmetry $\sin z=\sin(\pi-z)$.

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    $\begingroup$ good one (+1). We could also integrate $f(z)=\frac{e^{ixz}}{\sqrt{1-z^2}}$ around a big halfcircle in the uhp shifted infintesimaly away from the real axis. $\endgroup$ – tired Dec 3 '17 at 21:13
  • $\begingroup$ Thanks for this answer! $\endgroup$ – X. Sonorg Dec 4 '17 at 15:34

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