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This question arises from my answer to an inverse Laplace transform question. The result I got took the form

$$ f(t)= e^{-r_0 t/2} H(t-a) \left [ J_0\left(\frac{1}{2} a r_0\right) I_0\left(\frac{1}{2} r_0 t\right) \\+ 2 \sum_{k=1}^{\infty} J_{2 k}\left(\frac{1}{2} a r_0\right) I_{2 k}\left(\frac{1}{2} r_0 t\right)\right ] $$

where $H$ is the Heaviside step function:

$$H(x) = \begin{cases} \\ 1 & x > 0\\ 0 & x < 0\end{cases}$$

This result in turned derived from the following integral:

$$\frac{1}{\pi} e^{-r_0 t/2} \int_0^{\pi} d\theta \: \cos{\left(\frac{1}{2} a r_0 \sin{\theta}\right)} e^{(r_0 t/2) \cos{\theta}}$$

Now, I suspect that, if I could evaluate that sum in closed form, it would take the form

$$S\left(\sqrt{t^2-a^2}\right)$$

but, I stress, this is only a suspicion at this point. My question is two-fold: 1) is anyone aware of a closed form expression for that sum, and 2) even if not, is there a way to prove or disprove that the sum has the above functional form?

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Ron Gordon
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1 Answers1

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OK folks, I think I may have the answer to this question. Please correct me if I am wrong, but if I am right, then I looked at the original problem all wrong.

Consider the following integral:

$$\int_0^{\pi} d\theta \: \cos{\left(q \sin{\theta}\right)} e^{p \cos{\theta}} = \Re{\left[\int_0^{\pi} d\theta \: e^{i q \sin{\theta}} e^{p \cos{\theta}} \right]}$$

Now write

$$p=\sqrt{p^2-q^2} \cosh{\beta}$$ $$q=\sqrt{p^2-q^2} \sinh{\beta}$$

where $\beta = \text{arctanh}{(q/p)}$. The integral then becomes

$$\Re{\left[\int_0^{\pi} d\theta \: e^{\sqrt{p^2-q^2} \, \cos{(\theta + i \beta)}}\right]}$$

This integral is independent of $\beta$; to see this, differentiate with respect to $\beta$ and note that we are taking the real part of the integral. (And, for good measure, assume that $p \gt q$.) Therefore the integral in question is simply

$$\int_0^{\pi} d\theta \: \cos{\left(q \sin{\theta}\right)} e^{p \cos{\theta}} = \pi I_0\left(\sqrt{p^2-q^2}\right)$$

The solution to the problem is then

$$f(t) = H(t-a) I_0\left(\frac{1}{2} r_0 \sqrt{t^2-a^2}\right) e^{-r_0 t/2} $$

Ron Gordon
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