let $(\lambda_n)_{n \in N}$ be a sequence creasing of positive reals going to infinity .
$f(x)= \sum_{n=1}^{\infty}( -1)^ne^{- \lambda_n x}$
how can we calculate $\int_{0}^{\infty} f(t)dt$
for this , I wanna show that f is cintinuous on $I=[0,\infty[$ . so i ll show the continuity on all segments of I
for ech natural number the function $f_n(x)=(-1)^ne^{-\lambda_n x}$ is continuous on $I$, we will show the uniform convergence of $f_n$ in ech segment of I .
how can I process to show the continuity ? after that I think I can intervat thr sum with the integral !
$\int_{0}^{\infty} \sum_{n=1}^{\infty}( -1)^ne^{- \lambda_n t} dt$
= $\sum_{n=1}^{\infty} \int_{0}^{\infty} ( -1)^ne^{- \lambda_n t} dt$
$\int_{0}^{x} ( -1)^ne^{- \lambda_n t} dt$
=$ ( -1)^n \frac{-1}{\lambda_n}e^{- \lambda_n x}- (-1)^n$
its limit when x goes to infinity is $( -1)^n \frac{-1}{\lambda_n}+(-1)^{n+1}$
then we calculate $\sum_{n=1}^{\infty} ( -1)^{n+1}( \frac{1}{\lambda_n}+1)$
You made a mistake integrating, you should get $(-1)^n\frac{-1}{\lambda_n}e^{-\lambda_n x}-(-1)^n\left(\frac{-1}{\lambda_n}\right)$. Then, when taking $x\to\infty$ you get $\frac{(-1)^n}{\lambda_n}$. So, at the end your integral would be $\sum_{n=1}^\infty \frac{(-1)^n}{\lambda_n}$, which now makes sense (converges) because of the alternating series test.