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I need to evaluate the following integral(Inverse Hankel Transform of $\cos(k)$)

$$ f(r) = \int^{\infty}_{0} k\cos(k) J_{0}(kr) \space dk$$ where $$ J_{0}(x) $$ is the zeroth order Bessel function. Any help is highly appreciated.

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  • $\begingroup$ expand cosine function into power series and look on internet or something the value of the formulae $ \int _{0}^{\infty} k^{n}J_{0} (kr) $ $\endgroup$ – Jose Garcia Apr 26 '17 at 19:02
  • $\begingroup$ Does your integral converge at $\infty$? $\endgroup$ – DisintegratingByParts Apr 26 '17 at 19:05
  • $\begingroup$ $k\cos(k)\,J_0(kr)$ is not a $L^1(\mathbb{R}^+)$ function, since $J_0(kr)$ only decays like $\frac{1}{\sqrt{k}}$ for large $k$s. You have to use a suitable regularization for assigning a value to such divergent integral. $\endgroup$ – Jack D'Aurizio Apr 26 '17 at 19:10
  • $\begingroup$ @JackD'Aurizio : What is that suitable regularization? $\endgroup$ – DisintegratingByParts Apr 26 '17 at 23:05
  • $\begingroup$ @TrialAndError : a possibility is given by considering (if it is finite) $$\lim_{\varepsilon\to 0^+}\int_{0}^{+\infty}k\cos(k) J_0(kr) e^{-\varepsilon k}\,dk$$ $\endgroup$ – Jack D'Aurizio Apr 26 '17 at 23:17
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This integral does not converge.

The reason: The asymptotic expansion $J_0(k) \approx \sqrt{\tfrac{2}{\pi k}}\cos(k-\tfrac{\pi}{4})$

(see (http://www.nbi.dk/~polesen/borel/node15.html)) formula 153, gives asymptotically, up to a constant factor, an integrand of the form

$$\sqrt{k}cos(k)cos(k-\tfrac{\pi}{4})=\tfrac{\sqrt{k}}{2}\underbrace{(cos(2k-\tfrac{\pi}{4})+\cos(\tfrac{\pi}{4}))}_{f(k)}$$

$f$ is a periodic fonction with period $\pi$ with the following representative curve:

enter image description here

Let $a_m=(4m+2)\tfrac{\pi}{4}, b_m=(4m+3)\tfrac{\pi}{4}, c_m=(4m+4)\tfrac{\pi}{4}, a_m=(4m+5)\tfrac{\pi}{4}$. Then

$$I:=\int_0^{\infty}\sqrt{k}f(k)dk=\sum_{m=0}^{\infty}\left(\left(\underbrace{\int_{a_m}^{b_m}}_{I_m}+\underbrace{\int_{b_m}^{c_m}}_{J_m}+\underbrace{\int_{c_m}^{d_m}}_{K_m}+\underbrace{\int_{d_m}^{a_{m+1}}}_{L_m}\right)\sqrt{k}f(k)dk\right)$$

We admit here that it can be proven in a rigorous manner (taking into account the variable factor $\sqrt{k}$) that, for all $m$, $I_m+K_m>0$.

Thus $$I>\sum_{m=0}^{\infty}\left(\left(\underbrace{\int_{b_m}^{c_m}}_{J_m}+\underbrace{\int_{d_m}^{a_{m+1}}}_{L_m}\right)\sqrt{k}f(k)dk\right)$$

But the last sum contains an infinite number of times the same positive constant.

Thus this RHS is divergent, and therefore $I$ also.

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Mathematica gives this result, which is expressed in terms of the Meijer G-function: $$f(r)= 2 \sqrt{\pi}\, G_{1,1}^{0,1}\!\left(\frac{1}{r^2}\biggl| \begin{array}{c} 1 \\ \frac{3}{2} \\ \end{array} \right)\,.$$ In explicit form, this becomes $f(r)=\begin{cases} {-1\over(1-r^2)^{3/2}}\,,&\text{for $0<r<1\,,$}\\ 0\,,&\text{for $r>1\,.$}\end{cases}$

Additionally, the program package is able to produce a symbolic solution for the $\nu^{\text{th}}$-order inverse Hankel transforms of the cosine function: $$f_{\nu}(r)=\int_0^\infty k \cos(k) J_\nu(k r)\,dr= 2\sqrt\pi\, G_{2,2}^{1,1}\,\left(\frac{1}{r^2}\biggl| \begin{array}{c} 1-\frac{\nu }{2},1+\frac{\nu }{2} \\ 1,\frac{3}{2} \\ \end{array} \right).$$ This can be written in explicit form as $$f_{\nu}(r)=\begin{cases} \frac{-\cos \left(\frac{\pi \nu }{2}\right)\left(\frac{r}{\sqrt{1-r^2}+1} \right)^{\nu } \left(\nu\sqrt{1-r^2}+1\right)} {\left(1-r^2\right)^{3/2}}\,,&\text{for $0<r<1\,,$}\\ \frac{\nu \cos \left(\nu \csc ^{-1}(r)\right)}{r^2-1}+ \frac{\sin \left(\nu \csc^{-1}(r)\right)}{\left(r^2-1\right)^{3/2}}\,,& \text{for $r>1\,.$}\end{cases}$$

Disclaimer: I haven't done careful validity checks on these results. The Mathematica functions for symbolic computation of the distributional Hankel transform became available only upon the recent release of MMA Version 11.1.1.

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