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In the following, we try to calculate two integrations, \begin{eqnarray*} \int_{1}^{\infty}dy\sqrt{y^{2}-1}\cos(ay),\ \ \int_{1}^{\infty}dy\sqrt{y^{2}-1}\sin(ay),\ \ \text{with }a>0 \end{eqnarray*}

Mathematica tell me that these two integrations are not convergent. enter image description here

But I will use another method to calculate them. Firstly, I calculate following integration,

\begin{eqnarray*} \mathrm{I} & = & \int_{1}^{\infty}dy\sqrt{y^{2}-1}e^{-iay}=\int_{1}^{\infty}dy\sqrt{y^{2}-1}\cos(ay)-i\int_{1}^{\infty}dy\sqrt{y^{2}-1}\sin(ay) \end{eqnarray*}

We can see

\begin{eqnarray*} \mathrm{I} & = & \int_{1}^{\infty}dy\sqrt{y^{2}-1}e^{-iay}\\ & = & \int_{1}^{\infty}dy(y^{2}-1)\frac{1}{\sqrt{y^{2}-1}}e^{-iay}\\ & = & -(\frac{\partial^{2}}{\partial a^{2}}+1)\int_{1}^{\infty}dy\frac{e^{-iay}}{\sqrt{y^{2}-1}}\\ & = & -(\frac{\partial^{2}}{\partial a^{2}}+1)(-\frac{i\pi}{2})H_{0}^{(2)}(a)\\ & = & \frac{i\pi}{2}[H_{0}^{(2)\prime\prime}(a)+H_{0}^{(2)}(a)]\\ & = & \frac{i\pi}{2a}H_{1}^{(2)}(a)\\ & = & \frac{i\pi}{2a}[J_{1}(a)-iN_{1}(a)]\\ & = & \frac{\pi}{2a}N_{1}(a)+\frac{i\pi}{2a}J_{1}(a) \end{eqnarray*}

where we have used \begin{eqnarray*} \int_{1}^{\infty}dy\frac{e^{-iay}}{\sqrt{y^{2}-1}} & = & -\frac{i\pi}{2}H_{0}^{(2)}(a) \end{eqnarray*} \begin{eqnarray*} H_{0}^{(2)\prime\prime}(a)+H_{0}^{(2)}(a) & = & \frac{1}{a}H_{1}^{(2)}(a) \end{eqnarray*} \begin{eqnarray*} H_{n}^{(2)}(a) & = & J_{n}(a)-iN_{n}(a),\ n=0,1,2,\cdots \end{eqnarray*}

with $J_{n}(a)$, $N_{n}(a)$ and $H_{n}^{(2)}(a)$ the $n$-th Bessel funcion, $n$-th Neumann function and $n$-th Hankel function of the second kind. Finally, we get

\begin{eqnarray*} \int_{1}^{\infty}dy\sqrt{y^{2}-1}\cos(ay) & = & \frac{\pi}{2a}N_{1}(a)\\ \int_{1}^{\infty}dy\sqrt{y^{2}-1}\sin(ay) & = & -\frac{\pi}{2a}J_{1}(a) \end{eqnarray*}

Is my calculation right?

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    $\begingroup$ you can't exchange integration and differentiation in this case. nevertheless your identity might be valid in the sense of distributions $\endgroup$
    – tired
    Jul 19, 2016 at 9:31
  • $\begingroup$ @tired...Excuse me, my major is physics. What does "...valid in the sense of distributions " mean? $\endgroup$ Jul 19, 2016 at 11:54

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