Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

up vote 2 down vote accepted

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

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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.