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I've been examining an integral I remember encountering in complex analysis:

$$ I = \int_0^\pi e^{\cos\theta} \cos\left(n\theta - \sin\theta\right) d\theta $$

where $n$ is a nonnegative integer. It's not too hard to show that $I = \frac{\pi}{n!}$ by changing $\cos\left(n\theta - \sin\theta\right)$ into $\operatorname{Re}\left[e^{i(n\theta - \sin\theta)}\right]$ and making the substitution $z = e^{i\theta}$ to turn the integral into a contour integral on the complex unit circle.

However, I came across the following representation of the Bessel functions of the first kind on Wikipedia:

$$ J_n(x) = \frac{1}{\pi} \int_0^\pi \cos\left(n\theta - x\sin\theta\right) d\theta $$

Given the similarity between this formula when $x = 1$ and the initial integral, I'm sure there's some way to evaluate the integral using properties of the Bessel functions. But so far I'm stuck as to figuring it out. I tried expanding $e^{\cos\theta}$ as a series but that got ugly pretty quickly. Any suggestions?

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If we compute the Fourier cosine series of $e^{\cos\theta}$ over $[-\pi,\pi]$ we get: $$ e^{\cos\theta} = I_0(1)+2\sum_{k\geq 1}I_k(1)\cos(k\theta)\tag{1}$$ where $I_k$ is a modified Bessel function of the first kind. On the other hand, by using Briggs formulas and your definition of $J_n(x)$ it follows that: $$ \frac{1}{2\pi}\int_{-\pi}^{\pi}\cos(n\theta-\sin\theta)\,d\theta = J_n(1),$$ $$ \frac{1}{\pi}\int_{-\pi}^{\pi}\cos(n\theta-\sin\theta)\cos(k\theta)\,d\theta = J_{n+k}(1)+J_{n-k}(1),\tag{2}$$ so we have: $$ \int_{-\pi}^{\pi}e^{\cos\theta}\cos(n\theta-\sin\theta)\,d\theta =\\2\pi \cdot I_0(1)\,J_n(1)+2\pi\cdot\sum_{k\geq 1}I_k(1)\left(J_{n-k}(1)+J_{n+k}(1)\right)\tag{3}$$ and now we may exploit: $$ \sum_{n=-\infty}^{+\infty}t^n\,J_n(z) = \exp\left(\frac{z}{2}\left(t-\frac{1}{t}\right)\right) \tag{4}$$ together with $I_n(z)=(-i)^n\,J_n(iz)$ to write $(3)$ as a convolution and deduce: $$ I = \pi\cdot [z^n]\,e^{z} = \frac{\pi}{n!}\tag{5} $$ in a extremely convoluted and probably useless very nice way.

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  • $\begingroup$ Thanks! I'm still a little unclear on the convolution part. I understand up to where $I = \pi \sum_{k=-\infty}^\infty I_k(1) J_{n-k}(1) = \pi \sum_{k=-\infty}^\infty (-i)^k J_k(i) J_{n-k}(1)$ but I don't see how to apply eqn (4) to that to get eqn (5). $\endgroup$ – Billy R Jun 1 '15 at 17:42
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    $\begingroup$ @BillyR: just read the RHS of $(3)$ as the coefficient of $t^n$... in the right product. $\endgroup$ – Jack D'Aurizio Jun 1 '15 at 18:11
  • $\begingroup$ Ahh, I see it now. Thanks again! $\endgroup$ – Billy R Jun 1 '15 at 19:11

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