How do I evaluate $\int_{0}^{\infty}e^{-x}f(x)dx$. Suppose that a function $f(x)$ is defined as the sum of a series $f(x)=1-\frac{1}{(2!)^2}{(2015x)^2}+\frac{1}{(4!)^2}{(2015x)^4}-..........=\sum_{n=0}^{\infty}(-1)^n\frac{1}{(2n!)^2}{(2015x)^{2n}}$.
Evaluate $\int_{0}^{\infty}e^{-x}f(x)dx$.
My idea: Let $E \subset(0, \infty)$, and let $f_n(x)=(-1)^n\frac{1}{(2n!)^2}{(2015x)^{2n}}$, then
$|f_n(x)|=|(-1)^n\frac{1}{(2n!)^2}{(2015x)^{2n}}|\leq \frac{1}{(2n!)^2}{(2015x)^{2n}}$.
Let $M_n=\frac{1}{(2n!)^2}{(2015x)^{2n}}$. By Ratio test, $\sum M_n $ converges.
By W-M test $f_n$ converges uniformly. We know that if $f_n \rightarrow f$ converges uniformly then $f$ is continuous.
I know that if $f:(0,\infty) \rightarrow \mathbb{R}$, is continuous and bounded , then $\int_0^{\infty}e^{-x}f(x)=f(c)$ for some $c\in (0,\infty)$.
My question is how do I claim $f$ is bounded on $E$? Anyone can suggest me some hints?

Notice that $f_n$ converges to $f$ uniformly on any compact set, so for any $t > 0$ we have
$$ \int_0^t e^{-x} f(x) dx = \sum_{n=0}^\infty \int_0^t e^{-x} f_n(x) dx. $$
Further, we have
$$ \left| \int_0^t e^{-x} f_n(x) dx \right| \le \frac1{((2n)!)^2} \int_0^t e^{-x} (2015x)^{2n} dx \le \frac{(2n)!}{((2n)!)^2} \frac1{2015} = \frac{1}{(2n)!} \frac1{2015} =: M_n. $$
Finally, as $\sum_n M_n \le \frac e{2015} < \infty$, we have by Tannery's theoreom
$$ \lim_{t\to\infty} \sum_{n=0}^\infty \int_0^t e^{-x} f_n(x) dx = \sum_{n=0}^\infty \lim_{t\to\infty}\int_0^t e^{-x} f_n(x) dx = \sum_{n=0}^\infty \int_0^\infty e^{-x} f_n(x) dx, $$
which can be easily computed.
