I cannot find a closed form solution for the following integral:
$$\int_0^t\exp\left[- \frac{a(\delta+1)(t-x)}{(\delta-(t-x))}\right] \, b\exp[-bx] dx$$
So, to evaluate this integral, I have used numerical integration (in Matlab). Note that $a$, $b$, $\delta$ and $t$ are all positive. When I set $\delta I suspect that the problem results from the denominator $\delta-(t-x)$, which is equal to $0$ if $x= t-\delta$. My question: does the above mean that the integral diverges and goes to infinity for $\delta
Evaluate $\int_0^t\exp\left[- \frac{a(\delta+1)(t-x)}{(\delta-(t-x))}\right] \, b\exp[-bx] dx$
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integration
definite-integrals
numerical-methods
discontinuous-functions
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0what is the meaning of this brackets $$[...]$$? – 2017-02-21
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0$\exp[..]$ is equivalent to $\exp(..)$. Maybe it is a bad choice to use this notation. Please feel free to edit the question. – 2017-02-21
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0can you say something about the parameters? – 2017-02-21
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0This is not an integral unless you decide something for $d$, say $dx$? – 2017-02-21
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0Thank you for pointing out this issue. I have edited the question. – 2017-02-21
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0@Dr.SonnhardGraubner As indicated in the question, these parameters are all positive. $\delta$ can take values such that $\delta \ge t$ and also $\delta
1 Answers
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The first thing to do is to change variables $t-x=tz$, yielding
$$
I=\int_0^t\exp\left[- \frac{a(\delta+1)(t-x)}{(\delta-(t-x))}\right] \, b\exp[-bx]dx=bt e^{-bt}\int_0^1 dz \exp\left[- \frac{a(\delta+1)tz}{\delta-tz}\right] e^{btz}
$$
$$
\propto \int_0^1 dz \exp\left[- \frac{Az}{B-z}\right] e^{Cz}\ ,
$$
where $A$, $B$ and $C$ are the only combinations of parameters that are really relevant. Now, if $0
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0Just to make sure I understand your answer, for $\delta < t$ having $I$ equals to infinity is normal, meaning that the integral diverges if $\delta < t$, ? – 2017-02-21
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0Yes, that's correct – 2017-02-21