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In a dual integral situation, the following integral has to be involved $$ \int_0^\infty \frac{\cos (qt) J_1 (qr)}{1+q^2} \, \mathrm{d} q \quad\quad (0<t<r) \, . $$

Visibly this integral is convergent. I was wondering whether an amenable analytical expression is possible? This will be useful for my further analysis.

Any help is highly appreciated.

Thanks.

R

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    $\begingroup$ Don't know if it helps but solving this integral is equivalent to solve the differential equation $$ I''(t)-I(t)=\int_0^{\infty}\cos(q t)J_1(q r)dq $$ $\endgroup$ – tired Oct 5 '16 at 12:34
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    $\begingroup$ if you find something let me know, would be very interested $\endgroup$ – tired Oct 5 '16 at 12:43
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    $\begingroup$ I have done calculations following tired's idea and I got $$ \int_{0}^{\infty} \frac{\cos (tq) J_1(rq)}{1+q^2} \, dq = \frac{1}{r} - K_1(r) \cosh t \qquad 0 \leq t \leq r $$ where $K_n$ is the modified Bessel function of the 2nd kind. $\endgroup$ – Sangchul Lee Oct 5 '16 at 12:50
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    $\begingroup$ @SangchulLee the integral you mentioned is a special case of 10.22.59 here dlmf.nist.gov/10.22 $\endgroup$ – tired Oct 5 '16 at 13:06
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    $\begingroup$ @SangchulLee math.stackexchange.com/a/1829730 $\endgroup$ – Random Variable Oct 5 '16 at 15:18
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Following @tired's idea and using two known integral identities, we can compute the integral. Fix $r > 0$ and consider $I$ defined by

$$I(t) = \int_{0}^{\infty} \frac{\cos (tq) J_1 (rq)}{1+q^2} \, dq. $$

On the interval $(0, r)$, it satisfies the following 2nd ODE

$$ I(t) - I''(t) = \int_{0}^{\infty} \cos(tq)J_1(rq) \, dq, \qquad I(0) = \int_{0}^{\infty} \frac{J_1 (rq)}{1+q^2} \, dq, \quad I'(0) = 0. $$

We have two extra unknown integrals, but they can be computed using DLMF 10.22.59 and DLMF 10.22.46: for $0 < t < r$,

$$ \int_{0}^{\infty} \cos(tq)J_1(rq) \, dq = \frac{1}{r} \quad \text{and} \quad \int_{0}^{\infty} \frac{J_1 (rq)}{1+q^2} \, dq = \frac{1}{r} - K_1(r) \tag{*}$$

Thus the problem boils down to solving

$$ I(t) - I''(t) = \frac{1}{r}, \qquad I(0) = \frac{1}{r} - K_1(r), \quad I'(0) = 0. $$

Now the general solution of this equation is of the form

$$ I(t) = \frac{1}{r} + A \cosh t + B \sinh t $$

and plugging the initial condition shows $A = -1$ and $B = 0$. Therefore

$$ \int_{0}^{\infty} \frac{\cos (tq) J_1 (rq)}{1+q^2} \, dq = \frac{1}{r} - K_1(r) \cosh t, \qquad 0 < t < r. $$


p.s. I would love to see a self-contained solution as I don't quite understand $\text{(*)}$.

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    $\begingroup$ @tired, Absolutely. I don't understand why I was such in a hurry. $\endgroup$ – Sangchul Lee Oct 5 '16 at 13:41
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    $\begingroup$ For the integral resulting in the modified bessel function, show that $q^2I''(q)+qI'(q)=(q^2+1)I(q)$ which is the defining differential equation for this special function. this can be done by using the usual recurence realtions for bessel functions $\endgroup$ – tired Oct 5 '16 at 14:50
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    $\begingroup$ @tired, That makes sense to me. Thank you! I feel like I am really bad at special functions... :p $\endgroup$ – Sangchul Lee Oct 5 '16 at 14:57
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    $\begingroup$ Me too...but as soon as you not see them as random symbols but know something about their mathematical/physical background they become quiet interesting. So i try to get better ^^ $\endgroup$ – tired Oct 5 '16 at 14:59
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    $\begingroup$ Last comment: For second integral you want a proof for i think it is best to first calculate the laplace transform of $J_1$ which should be not to difficult using the series representation. afterwards one can perform a proper analytic continuation which (should) yield the correct result $\endgroup$ – tired Oct 5 '16 at 15:14

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