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$$ \int_0^ \infty e^{-x\sinh(t)-\frac{1}{2}t}~dt $$ I tried doing it by parts and looking for differentials but I just keep getting back to the original expression. I can't think of a clever substitution either. Mathematica is giving me a complex answer with special functions: $$ \frac{e^{-ix}\sqrt{\frac{\pi}{2}}(-i+ie^{2ix}\text{Erfc}[(-1)^{\frac{1}{4}}\sqrt{x}]+\text{Erfi}[(-1)^\frac{1}{4}\sqrt{x}])}{\sqrt{x}} $$ For real $x>0$, it does evaluate to real answers though.

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$\int_0^\infty e^{-x\sinh t-\frac{t}{2}}~dt$

$=2\int_\infty^0e^\frac{x(e^{-t}-e^t)}{2}~d(e^{-\frac{t}{2}})$

$=2\int_0^1e^{\frac{xt^2}{2}-\frac{x}{2t^2}}~dt$

$=\left[\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{i\sqrt xt}{\sqrt2}+\dfrac{\sqrt x}{\sqrt2t}\right)+e^{-ix}~\text{erfc}\left(\dfrac{i\sqrt xt}{\sqrt2}-\dfrac{\sqrt x}{\sqrt2t}\right)\right)\right]_0^1$ (according to http://dlmf.nist.gov/7.7#E7)

$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}\left(\text{erfc}\left(\dfrac{(i+1)\sqrt x}{\sqrt2}\right)-\text{erfc}(+\infty)\right)+e^{-ix}\left(\text{erfc}\left(\dfrac{(i-1)\sqrt x}{\sqrt2}\right)-\text{erfc}(-\infty)\right)\right)$

$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{(i+1)\sqrt x}{\sqrt2}\right)+e^{-ix}\left(\text{erfc}\left(\dfrac{(i-1)\sqrt x}{\sqrt2}\right)-2\right)\right)$

$=\dfrac{i\sqrt\pi}{\sqrt{2x}}\left(e^{ix}~\text{erfc}\left(\dfrac{(1+i)\sqrt x}{\sqrt2}\right)-e^{-ix}~\text{erfc}\left(\dfrac{(1-i)\sqrt x}{\sqrt2}\right)\right)$

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  • $\begingroup$ Gods, thanks, this seems like what I'm looking for. I don't know how you managed to scrounge up the motivation to actually look it up in that ocean of a handbook, but thanks a lot. $\endgroup$ – Kurome May 8 '14 at 4:02
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You can have a closed form in terms of the AngerJ function

$$ \frac{2}{x}+\sqrt{2\pi}\,\frac{\cos(x)}{x^{3/2}}+\sqrt{2\pi}\frac{\sin(x)}{x^{1/2}}-\pi {\bf{J}}_{\frac{1}{2}} ( x )-\frac{\pi}{x}{ \bf{J}}_{-\frac{1}{2}} ( x ) .$$

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  • $\begingroup$ How were you able to express the integral in terms of the Anger function? I know the integral is actually part of one term in the integral representation of Bessel function $J_{1/2}(x)$. $\endgroup$ – Kurome Apr 16 '14 at 3:40
  • $\begingroup$ @Mhenni Do you know how to do the power series? $\endgroup$ – Jeff Faraci Apr 16 '14 at 4:06
  • $\begingroup$ @Integrals: Do you mean using power series technique to evaluate the integral? $\endgroup$ – Mhenni Benghorbal Apr 16 '14 at 4:12
  • $\begingroup$ @MhenniBenghorbal Yes earlier you commented on the post and asked about power series solution, I am still looking. THanks $\endgroup$ – Jeff Faraci Apr 16 '14 at 4:45
  • $\begingroup$ Do you thinj it's possible to put it into a closed form that does not contain Anger or Bessel functions? $\endgroup$ – Kurome Apr 16 '14 at 11:28

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