Integration of power series I am trying to find an approximate solution of:
\begin{equation}
\int_{0}^{+\infty} e^{-x}dx \ (=1)
\end{equation}
from the power series expansion of :
\begin{equation}
e^{-x}= 1-x+(1/2)\cdot x^2-(1/6)\cdot x^3+(1/24)\cdot x^4-(1/120)\cdot x^5+O(x^6)
\end{equation}
My Problem is that when I integrate the series term by term, the polynomials wont behave well with the $\infty$ term..
Can you help me please
 A: You have to remember that the polynomials in power series of nonpolynomial functions are infinite.
And if you have an infinite bound, the polynomial will grow unboundedly without all the other infinity terms to balance out the polynomial and converge it to $e^{-x}$.
Also how were you planning to plug in infinity? 
A: An alternative approach is first making a change of variable in your integral such that your improper integral becomes a proper integral, and use the power series expansion of the new integral to approximate your result.
By example with the change of variable $-x=\ln t$ we get
$$\int_0^\infty e^{-x}\mathrm dx=\int_0^1\mathrm dt=t\Big|_0^1$$
what is trivial in this case because the function $f(x)=x$ is identical to it power expansion.
But as the answers shown above we cannot use a power expansion to approximate an integral when one of it limits is infinity because for any non constant polynomial $p$ we have that
$$\lim_{x\to\infty}p(x)\in\{-\infty,\infty\}$$

As a complementary comment: another problem to approximate an integral using power expansions is that, in many cases, we have that the range of integration is not included in the radius of convergence, so we first need rewrite the integral in a finite/infinite sum of ranges of integration and use a different power expansion adapted to each range of integration.
A: The improper integral $\int_0^\infty e^{-x}\,dx$ is equal to 
\begin{align*}
    \lim_{t\to\infty} \int_0^t e^{-x}\,dx 
    &= \lim_{t\to\infty} \int_0^t \sum_{k=0}^\infty \frac{(-1)^k x^k}{k!}\,dx \\
    &= \lim_{t\to\infty} \int_0^t \lim_{n\to\infty} \sum_{k=0}^n \frac{(-1)^k x^k}{k!}\,dx \\
    &= \lim_{t\to\infty} \lim_{n\to\infty} \int_0^t \sum_{k=0}^n \frac{(-1)^k x^k}{k!}\,dx \\
    &= \lim_{t\to\infty} \lim_{n\to\infty} \left. \sum_{k=0}^n \frac{(-1)^k x^{k+1}}{(k+1)!}\right|^t_0 \\
    &= \lim_{t\to\infty} \lim_{n\to\infty} \sum_{k=1}^n \frac{(-1)^{k+1} t^{k}}{k!} \\
\end{align*}
What you are trying to do is swap the $\lim_{t\to\infty}$ and $\lim_{n\to\infty}$ operations.  But this doesn't make any sense, because for a fixed $n$,
$$
    \lim_{t\to\infty} \sum_{k=1}^n \frac{(-1)^{k+1} t^{k}}{k!} = 
    \begin{cases}
        +\infty & \text{if $n$ is odd} \\
        -\infty & \text{if $n$ is even}
    \end{cases}
$$
It's always dangerous to cavalierly interchange infinite processes.  If it's allowed, it's because of a theorem with certain hypotheses, and there are counterexamples when those hypotheses aren't met.  
