Laplace transform of $( \sin t)/t$ 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 $s&gt;0$ 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.
The interchange of sum and integral is permissible if $\sum \int |\frac {e^{-st}t^{2k}} {(2k+1)!}|\, dt &lt;\infty $ [This is by Fubini's Theorem]. For $t &lt;1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \sum \frac {e^{-st}} {(2k+1)!} &lt; e^{1-st}$. For $t &gt;1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \frac 1 t \sum \frac {t^{2k+1}e^{-st}} {(2k+1)!} &lt; e^{-(s-1)t}$. 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}) &lt;\infty $  for $s &gt;1$.
The interchange of sum and integral is permissible if $\sum \int |\frac {e^{-st}t^{2k}} {(2k+1)!}|\, dt &lt;\infty $ [This is by Fubini's Theorem]. For $t &lt;1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \sum \frac {e^{-st}} {(2k+1)!} &lt; e^{1-st}$. For $t &gt;1$, $\sum \frac {e^{-st}t^{2k}} {(2k+1)!} \leq \frac 1 t \sum \frac {t^{2k+1}e^{-st}} {(2k+1)!} &lt; e^{-(s-1)t}$. 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}) &lt;\infty $  for $s &gt;1$.