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I would like to calculate the following integral:

$$\int \exp\left[a \frac{1-e^{-\kappa_1 s}}{\kappa_1}+b\frac{1-e^{-\kappa_2 s}}{\kappa_2}+c\times s\right]ds$$

This is how I proceeded:

Let's define $u\triangleq e^{c\times s}$! Then we have:

$du=c\times u\times ds$

where $a, b, c, \kappa_1$ and $\kappa_2$ are constants.

Thus the original integral is:

$\int u \exp\left[\frac{a}{\kappa_1}+\frac{b}{\kappa_2}\right]\exp\left[-\frac{a}{\kappa_1}e^{-\kappa_1 s}-\frac{b}{\kappa_2}e^{-\kappa_2 s}\right]ds \\ =\frac{1}{c}\exp\left[\frac{a}{\kappa_1}+\frac{b}{\kappa_2}\right]\int \exp\left[-\frac{a}{\kappa_1}u^{-\frac{\kappa_1}{c}}-\frac{b}{\kappa_2}u^{-\frac{\kappa_2}{c}}\right]du$

But from here I could not go any further. Any hints and help would be greatly appreciated!

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  • $\begingroup$ Have you any reason to think that a closed form could exist for the antiderivative ? I guess that, for a small range of integration, we could approximate it. If this is the case, just post. $\endgroup$ – Claude Leibovici Jun 7 '15 at 2:15
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$$ \begin{align} \int e^{a\frac{1-e^{-\kappa_1s}}{\kappa_1}+b\frac{1-e^{-\kappa_2s}}{\kappa_2}+cs}\ \text{d}s & = e^{\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}\int e^{-\frac{ae^{-\kappa_1s}}{\kappa_1}-\frac{be^{-\kappa_2s}}{\kappa_2}}e^{cs}\ \text{d}s \\ & = e^{\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\left(\dfrac{ae^{-\kappa_1s}}{\kappa_1}+\dfrac{be^{-\kappa_2s}}{\kappa_2}\right)^ne^{cs}}{n!}\ \text{d}s \\ & = e^{\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^nC_k^n\dfrac{a^ke^{-\kappa_1ks}b^{n-k}e^{-\kappa_2(n-k)s}}{\kappa_1^k\kappa_2^{n-k}}e^{cs}}{n!}\ \text{d}s \\ & = e^{\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}\int\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^kb^{n-k}e^{(c-\kappa_1k-\kappa_2(n-k))s}}{k!(n-k)!\kappa_1^k\kappa_2^{n-k}}\ \text{d}s \\ & = \sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^na^kb^{n-k}e^{(c-\kappa_1k-\kappa_2(n-k))s+\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}}{k!(n-k)!\kappa_1^k\kappa_2^{n-k}(c-\kappa_1k-\kappa_2(n-k))}+C \\ & = \sum\limits_{k=0}^\infty\sum\limits_{n=k}^\infty\dfrac{(-1)^na^kb^{n-k}e^{(c-\kappa_1k-\kappa_2(n-k))s+\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}}{k!(n-k)!\kappa_1^k\kappa_2^{n-k}(c-\kappa_1k-\kappa_2(n-k))}+C \\ & = \sum\limits_{k=0}^\infty\sum\limits_{n=0}^\infty\dfrac{(-1)^{n+k}a^kb^ne^{(c-\kappa_1k-\kappa_2n)s+\frac{a}{\kappa_1}+\frac{b}{\kappa_2}}}{k!n!\kappa_1^k\kappa_2^n(c-\kappa_1k-\kappa_2n)}+C \end{align} $$

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  • 2
    $\begingroup$ Who says that one may commute the series with the indefinite integral? And, in general, what is one supposed to do with the formula you provide? Is that supposed to be a useful answer? $\endgroup$ – Alex M. Jun 12 '15 at 16:25

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