7

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, ...?

X. Sonorg
  • 93

2 Answers2

8

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 $\sin z=\sin(\pi-z)$.

Random Variable
  • 42,026
Paul Enta
  • 14,113
3

Alternatively, we can use a quarter circle.

Using the branch of the logarithm where $0\le \arg(z) < 2 \pi$, the function $$f(z) = \frac{\exp(ixz)}{\sqrt{z^{2}-1}} \, , \quad x>0, $$ has branch cuts on $(-\infty, -1]$ and $[1, \infty)$.

Let's integrate $f(z)$ around the following contour:

enter image description here

Jordan's lemma says that integral vanishes along the large quarter circle as $ R \to \infty$.

And since $ \lim_{z \to 1}(z-1)f(z) = 0$, the integral vanishes on the small semicircle around $z=1$ as the semicircle's radius goes to zero.

We therefore have $$ \int_{0}^{1} \frac{\exp(ixt)}{e^{i \pi /2}\sqrt{1-t^{2}}} \, \mathrm dt + \int_{1}^{\infty} \frac{\exp(ixt)}{\sqrt{t^{2}-1}} \, \mathrm dt - \int_{0}^{\infty} \frac{\exp(-xt)}{e^{i \pi/2} \sqrt{t^{2}+1}} \, e^{i \pi /2} \, \mathrm dt=0 \tag{1}. $$

And equating the imaginary parts on both sides of equation $(1)$, we have $$\int_{1}^{\infty} \frac{\sin(xt)}{\sqrt{t^{2}-1}} \, \mathrm dt = \int_{0}^{1} \frac{\cos(xt)}{\sqrt{1-t^{2}}} \, \mathrm dt = \int_{0}^{\pi/2} \cos(x \sin \varphi) \mathrm \, d \varphi = \frac{1}{2} \int_{0}^{\pi}\cos(x \sin \varphi) \, \mathrm d \varphi. $$

If we equate the real parts on both sides of equation $(1)$, we have

$$ \begin{align} \int_{1}^{\infty} \frac{\cos(xt)}{\sqrt{t^{2}-1}} \, \mathrm dt &= - \int_{0}^{1} \frac{\sin(xt)}{\sqrt{1-t^{2}}} \, \mathrm dt + \int_{0}^{\infty} \frac{\exp(-xt)}{\sqrt{t^{2}+1}} \, \mathrm dt \\ &\overset{(1)}{=} -\frac{\pi}{2} \, {\bf H}_{0}(x) + \frac{\pi}{2} \, {\bf K}_{0}(x) \\ &= -\frac{\pi}{2} \, {\bf H}_{0}(x) + \frac{\pi}{2} \left({\bf H}_{0}(x)-Y_{0}(x) \right) \\ &= -\frac{\pi}{2} \, Y_{0}(x). \end{align}$$

$(1)$ https://en.wikipedia.org/wiki/Struve_function#Integral_form

Random Variable
  • 42,026