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Is there some way to express:

$$I(t) = \int_{-\infty}^{t} e^{-2\mathrm{cosh}(x)}~\mathrm{d}x$$

From Bessel functions?

By substituting $y = \mathrm{cosh}(x)$ we get

$$I(t) = \int_{1}^{\mathrm{cosh}(t)} \frac{e^{-2 y}}{\sqrt{y^2-1}}~\mathrm{d}y$$

In this form, Mathematica will give $I(\infty) = \mathrm{BesseK}(0,2)$ but not indication of what $I(t)$ might be.

I can write the integral as:

$$I(t) = \int_{1}^{\mathrm{sinh}(t)} \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}~\mathrm{d}z$$

and try to expand $f(x) = \frac{e^{-2 \sqrt{z^2+1}}}{\sqrt{z^2+1}}$ as

$$f(z) = \sum_{i=0}^{\infty} (-1)^j\frac{2~\mathrm{BesselK}\left(\frac{1}{2}+j,2\right)}{j!\sqrt{\pi}} z^{2j}$$

Unfortunately the series diverges for $z \ge 1$, as does the series for $f(z) z e^{2 z}$

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    $\begingroup$ Are you sure that your integral is converge ? $\endgroup$ – zeraoulia rafik Feb 6 '17 at 22:28
  • $\begingroup$ Yes, it's positive and it converges to BesselK(0,2) for t = infinity. $\endgroup$ – Arthur B Feb 7 '17 at 0:19
  • $\begingroup$ Your $I(t)$ is an incomplete Bessel function, also known as leaky aquifer function. See papers by Jones (Proc. Edimb. Math. Soc. 2007), Fripiat and Harris and many others. Several expansions exist, depending on the range of the parameter $t$. You may find expansion in series of Bessel functions. $\endgroup$ – Paul Enta Feb 14 '17 at 19:02
  • $\begingroup$ See for example eq. (38) in Harris. $\endgroup$ – Paul Enta Feb 14 '17 at 20:26
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By changing $s=e^{t-x}$ in the integral, one obtains $$I(t)=\int_1^\infty e^{-e^{-t}s-e^t/s}\frac{ds}{s}$$ which corresponds to the definition of the incomplete Bessel function in Harris: $$ I(t)=K_0(e^{-t},e^t)$$ With eq. (38) in the article, $$I(t)=K_0(2)+tI_0(2)+2\sum_{j=1}^\infty \frac{(-1)^j}{j}I_j(2)\sinh(jt)$$ where $I_n$ and $K_n$ are the modified Bessel functions. The author mentions that the expansion is absolutely convergent.

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  • $\begingroup$ It does seem to converge, but empirically convergence gets very bad when $t$ is far from 0 as the $\mathrm{sinh}$ terms grow exponentially. I'm surprised that it's absolutely convergent $\endgroup$ – Arthur B Feb 16 '17 at 0:42

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