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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

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    $\begingroup$ This is something you don't wan to do. The problem: your interval is unbounded. $\endgroup$ – ncmathsadist Jun 6 '17 at 18:41
  • $\begingroup$ what means the notation $(=1)$? First time I see it. $\endgroup$ – Masacroso Jun 6 '17 at 18:50
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    $\begingroup$ @Masacroso I mean that i know the integral is 1, but willing to get it from the power series $\endgroup$ – outlawoutlawz Jun 6 '17 at 18:52
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    $\begingroup$ From the power series it is simple to check that $\frac{d}{dx}\left(1-e^{-x}\right)=e^{-x}$ and by the fundamental theorem of Calculus $\int_{0}^{+\infty}e^{-x}\,dx = \left[1-e^{-x}\right]^{+\infty}_0 = 1. $ However you cannot exchange $\sum_{n\geq 0}$ and $\int_{0}^{+\infty}$ as explained below. $\endgroup$ – Jack D'Aurizio Jun 6 '17 at 19:31
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    $\begingroup$ The kind of approximation you're using is valid in a neighborhood of 0 $\endgroup$ – Andrea Corbellini Jun 6 '17 at 21:34
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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.

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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?

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  • $\begingroup$ I wanted to use simple math as in integrating x from 0 to infinity for example, but i know this wont work, so i am asking if there is a way to go around it $\endgroup$ – outlawoutlawz Jun 6 '17 at 18:46
  • $\begingroup$ not if you have an infinity, sorry. and if the upper bound was 500, say, you'd still have FAR too few terms to do any close computation $\endgroup$ – Saketh Malyala Jun 6 '17 at 18:47
  • $\begingroup$ In theory it could be done like standard improper integration: $\displaystyle \int_0^{+\infty} = \lim_{A \to +\infty} \int_0^A$. The problem is simplifying the power series expansion for large (or any) values of $A$. I think this simplification is what can't be done. $\endgroup$ – tilper Jun 6 '17 at 19:02
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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.

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    $\begingroup$ +1 for raising the point that the range of integration might be beyond the interval of convergence for a power series. Infinity is not necessarily even the problem with this approach. $\endgroup$ – NoseKnowsAll Jun 6 '17 at 23:16

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