Is there anyway to evaluate the following integral in form of special function or maybe an infinite sum ?
$I =\int\limits_0^\infty {\frac{{{e^{ – x – \frac{1}{x}}}}}{{1 + x}}dx}$
Thank you very much !
Evaluate $\int\limits_0^\infty {\frac{{{e^{ – x – \frac{1}{x}}}}}{{1 + x}}dx}$ in form of special function
improper-integralsintegrationspecial functions
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Best Answer
We have, by the change of varaiables $x=e^t$, \begin{align*} \int_0^{ + \infty } {\frac{{e^{ - x - \frac{1}{x}} }}{{1 + x}}dx} & = \int_{ - \infty }^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{1 + e^t }}e^t dt} = \int_{ - \infty }^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt} \\ & = \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt} + \int_{ - \infty }^0 {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt} \\ & = \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{t/2} }}{2}dt} + \int_0^{ + \infty } {\frac{{e^{ - 2\cosh t} }}{{\cosh (t/2)}}\frac{{e^{ - t/2} }}{2}dt} \\ & = \int_0^{ + \infty } {e^{ - 2\cosh t} dt} = K_0 (2)=0.113893872\ldots, \end{align*} where $K_\nu(z)$ is the modified Bessel function of the second kind (cf. http://dlmf.nist.gov/10.32.E9).