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Could someone evaluate this integral for me?

$$ \int_0^\infty \int_0^\infty \left(u-\frac{u^2}{2\gamma} \right)\mathrm e^{-u/\gamma}\mathrm e^{-|t-u+v|/\mu}\left(v-\frac{v^2}{2\gamma} \right) \mathrm e^{-v/\gamma} \mathrm du\mathrm dv$$

Please note the absolute value in the exponent!

It would be of tremendous help! I've been struggling with this for so long!

Thank you!!

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  • $\begingroup$ What assumptions (if any) are there on the two parameters $t$ and $\gamma$? $\endgroup$ – David H May 21 '18 at 11:04
  • $\begingroup$ t, $\gamma$ and $\mu$ are positive reals. Thank you! $\endgroup$ – adamG May 21 '18 at 11:32
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Set $$ f(x)=\left(x-\frac{x^2}{2c}\right)e^{-x/c}\textrm{, }c>0, $$ then $$ I=\int^{\infty}_{0}\int^{\infty}_{0}\left(u-\frac{u^2}{2c}\right)e^{-u/c}e^{-|u-v-t|/m}\left(v-\frac{v^2}{2c}\right)e^{-v/c}dudv= $$ $$ =\int^{\infty}_{0}\int^{\infty}_{0}f(u)f(v)e^{-|u-v-t|/m}dudv= $$ $$ =\int^{\infty}_{0}\int_{0\leq u\leq v+t}f(u)f(v)e^{(u-v-t)/m}dudv+ $$ $$ +\int^{\infty}_{0}\int_{u\geq v+t}f(u)f(v)e^{-(u-v-t)/m}dudv= $$ $$ =\int^{\infty}_{0}\int^{v+t}_{0}f(u)f(v)e^{(u-v-t)/m}dudv+ $$ $$ +\int^{\infty}_{0}\int^{\infty}_{v+t}f(u)f(v)e^{-(u-v-t)/m}dudv= $$ $$ =e^{-t/m}\int^{\infty}_{0}f(v)e^{-v/m}\left(\int^{v+t}_{0}f(u)e^{u/m}du\right)dv+ $$ $$ +e^{t/m}\int^{\infty}_{0}f(v)e^{v/m}\left(\int^{\infty}_{v+t}f(u)e^{-u/m}du\right)dv. \tag 1 $$ But $$ f(u)e^{\pm u/m}=u\left(1-\frac{u}{2c}\right)\exp\left(-\left(\frac{1}{c}\pm \frac{1}{m}\right)u\right) $$ and if $Re\left(1/c+1/m\right)>0$ and $Re\left(1/c\right)>0$, then the integrals in (1) converge and we have $$ \int e^{ax}(kx+l)dx=e^{ax}\left(\frac{kx}{a}+\frac{la-k}{a^2}\right) $$ and $$ \int e^{ax}(kx^2+lx+m)dx=e^{ax}\left(\frac{kx^2}{a}+\frac{al-2k}{a^2}x+\frac{2k-al+a^2m}{a^3}\right). $$ Hence after elementary evaluations in (1) we get $$ I=\int^{\infty}_{0}\int^{\infty}_{0}\left(u-\frac{u^2}{2c}\right)e^{-u/c}e^{-|u-v-t|/m}\left(v-\frac{v^2}{2c}\right)e^{-v/c}dudv= $$ $$ =\frac{c^3me^{-t/c}}{8(c^2-m^2)^3}\left(c^6-6 c^4m^2-3 c^2 m^4+c(c^4+2c^2m^2-3m^4)t-(c^2-m^2)^2t^2\right)+ $$ $$ +\frac{c^6m^4e^{-t/m}}{(c^2-m^2)^3}. $$

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You can transform this into a do-able integral by making the substitution $x=t-u-v$ and $y=v$. The region of integration becomes $0<y<t-x$ and $-\infty<x<t$.

This makes the absolute value easier to handle since it is now just $|x|$. We can split up the interval $-\infty<x<t$ into $-\infty<x<0$, where $|x|=-x$, and $0<x<t$, where $|x|=x$.

Also, the Jacobian for this subtitution is $-1$.

The result of this transformation is the messy but do-able integral:

$$-e^{\frac{-t}{\gamma}}\left[\int\limits_{-\infty}^0\int\limits_0^{t-x}\left( t-x-y-\frac{1}{2\gamma}(t-x-y)^2 \right)e^{\frac{x}{\gamma}}e^{\frac{y}{\gamma}}e^{\frac{x}{\mu}}(y-\frac{y^2}{2\gamma})dydx\\+\int\limits_0^t\int\limits_0^{t-x}\left( t-x-y-\frac{1}{2\gamma}(t-x-y)^2 \right)e^{\frac{x}{\gamma}}e^{\frac{y}{\gamma}}e^{\frac{-x}{\mu}}(y-\frac{y^2}{2\gamma})dydx\right].$$

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