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I am trying to integrate this function:

$$I = \int_{-\infty }^0 \frac{e^u}{\left(\left(e^u (u-1)+1\right) \lambda -1\right){}^2} \, du$$

and I cannot do it.

So, what I want to try to do is to expand the numerator in its Taylor series and integrate term-by-term, I.e.,

$$I = \int_{-\infty }^0 \frac{1+u+\frac{u^2}{2}+\frac{u^3}{6}+\cdots}{\left(\left(e^u (u-1)+1\right) \lambda _R-1\right){}^2} \, du$$

$$= \int_{-\infty }^0 \frac{1}{\left(\left(e^u (u-1)+1\right) \lambda -1\right){}^2} \, du+\int_{-\infty }^0 \frac{u}{\left(\left(e^u (u-1)+1\right) \lambda-1\right){}^2} \, du\ + \cdots$$,

but each term individually diverges.

NOTE: $I$ does not diverge when $\lambda = 0.5$ - the value of the integral is 1.42537

How can this be?

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That's like taking the integral of $e^x$ from $(-\infty,0)$ by using its Maclaurin expansion... It's just not going to work in general unless you justify the interchangeability of the series and integral. To put it short, you have three limits: the lower bound is a limit, the integral is a limit of a Riemann sum (by definition), and the series expansion is also a limit (a limit of partial sums). In general, limits are not interchangeable...

For conditions on when you can do this, look up uniform convergence and the dominated convergence theorem.

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You're exchanging two limiting operations (integral and infinite sum), which requires conditions on the type of convergence of the sum for the integrand in order to show the exchange is true, such as uniform convergence.

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