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I want to integrate

$$\int_{0}^{t}\exp\left\{{k_{1}\left ( 1-e^{-t/{k_{2}}} \right )}\right\}dt$$

First I substituted $u = 1-e^{-t/{k_{2}}}$

Thus I get $$k_{2}\int_{0}^{t}\frac{e^{k_{1}u}}{1-u}du$$

Now I chose $w(u) = \frac{1}{1-u}$ and $x(u) = {e^{k_{1}u}}$

[$w(u)$ is $u(x)$ and $x(u)$ is $v(x)$ in integration by parts formula.]

But the solution yields another integral with a $\frac{1}{(1-u)^2}$ term as I have to take differential of w(u) for integration by parts.

How can I proceed? If possible I want a solution that doesn't involve an integral.

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  • $\begingroup$ If I may : in the first integral, change the upper bound to something else than "t". In the second integral, change the upper bound for the corresponding value of "u". $\endgroup$ Jan 11, 2014 at 5:25

2 Answers 2

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Your second integral is an exponential integral -- there is no closed form integral in terms of elementary functions.

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As Igor Rivin mentioned, your second integral can not be expressed using simple functions.

However, you could expand 1/(1-u) as a Taylor series built around u=0. This gives you the sum of powers u^n (from n=0 to infinity). As a result, you can then write the antiderivative corresponding to your second integral as

k2 Sum[R(n),n=1,Infinity]

with R(n) = Integral_of[ u^n Exp[k1 u] du ]

I would suggest a new change of variable z = k1 u which, omiting the leading coefficients, will lead to a weighted infinite sum of integrals of z^n Exp[z] dz.

Integrating once by parts, you will easily establish a recurrence relation for these last integrals.

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