I have a question about this example from Fundamentals of Differential Equations by Nagle:
Why can we interchange the infinite sum and the Laplace transform $\mathcal{L}$? Specifically, why is it true that
$$\int_0^\infty e^{-st}\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}t^{2k}dt=
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\int_0^\infty e^{-st}t^{2k}dt$$
I was thinking we must look at the infinite series
$$\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}e^{-st}t^{2k}$$
(with fixed $s>0$ and as a function of $t$). I was thinking either to a) show the partial sums of the above converge uniformly to swap the integral and infinite sum, but I think uniform convergence is false because the maximal value of each term of the sum occurs at $t=2k$, and we have a $(2k)^{2k}$ term that overpowers everything else.
or b) try to use Dominated convergence, but I need an integrable function that is an upper bound of each partial sum, which I'm also having trouble finding.
Thanks so much in advance for any help.
[Math] Laplace transform of $( \sin t)/t$
calculusintegrationlaplace transform
Best Answer
The interchange of sum and integral is permissible if $\sum \int |\frac {e^{-st}t^{2k}} {(2k+1)!}|\, dt <\infty $ [This is by Fubini's Theorem]. For $t <1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \sum \frac {e^{-st}} {(2k+1)!} < e^{1-st}$. For $t >1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \frac 1 t \sum \frac {t^{2k+1}e^{-st}} {(2k+1)!} < e^{-(s-1)t}$. Hence $e^{-(s-1)t}+e^{1-st}$ is an upper bound for the sum for all $t$. Of course $\int_0^{\infty } (e^{-(s-1)t}+e^{1-st}) <\infty $ for $s >1$.