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It's easy to show with integration by parts that for any n, $\int_0^\infty e^{-x}x^n dx$ is n!

Also for any function that can be represented equivalently as a taylor series around $x=0$, can be written in the form $f(x) = \sum_{n=0}^\infty {f^{(n)}(0)\ x^n\over n!}$.

So for any of these functions, $\int_0^\infty e^{-x}f(x)\ dx = \int_0^\infty e^{-x} \sum_{n=0}^\infty {f^{(n)}(0)\ x^n\over n!}\ dx$, reordering the sum and integral, we get $ \sum_{n=0}^\infty {f^{(n)}(0)\over n!} \int_0^\infty e^{-x}x^n\ dx $, and using the above identity, this simplifies to $\sum_{n=0}^\infty f^{(n)}(0)$

now let $f(x)=e^{-x}$, which follows all requirements to be expanded in a taylor expansion, and so from above $\int_0^\infty e^{-x}*e^{-x}\ dx $ is $\sum_{n=0}^\infty {d^n\over dx}e^{-x}|_0$ or simply $1-1+1-1+1......$

Although this integral can be straight forwardly evaluated as .5

So now this means $1+1-1+1-1+1... = .5$???

So where is the mistake? I know there have been other proofs showing this result but those fault somewhere in the limits taken. So here is the issue that in the limit as n$\rightarrow \infty$ for the n! identity?

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    You didn't take into account radius of covergence. $f(x)=\frac 1 {1-x}=1+x+x^2+x^3+... $ so $f (-1)=.5 =1-1+1-1+... $ which would be fine except x=-1 is right not in the radius of convergence. – 2017-02-12
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    How is the argument ''simply $1-1+1-1+1-\cdots$" made? And surely one needs a careful definition for the infinite sum. – 2017-02-12
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    How do you justify interchanging the series with the integral? Apparently, you took that very critical step for granted. – 2017-02-12
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    @fleablood The radius of convergence isn't the issue. The issue is the lack of uniform convergence. – 2017-02-12
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    Fair nuff. But point is the solution for cases with convergence work and consistant will yield correct results. Convergence does not apply. Sum = 1/2 should appear wrong, but not absurd. Surprised me first time I saw it. – 2017-02-12

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Your series is often called Grandi's Series.

You didn't make any mistakes except for trying to apply rules for uniformly convergent series to your series, which yields a known result that has meaning in some contexts but does not convergence in the usual sense.

On the Wikipedia page I linked you will find a number of ways to derive a result of $1/2$, but perhaps the simplest is to take the limit of $\frac{s_n}{n}$ as $n$ goes to infinity, where $s_n$ is the $n$th partial sum of Grandi's Series. In a specific sense, $1/2$ is what the series would be if it were convergent

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    "There's no error in your work." Yes,.there apparently is a very significant one. How on earth does one justify interchanging the series and integral? – 2017-02-12
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    @Dr.MV OK, let me clarify that sentence to be a but more explicit: "you didn't make any mistake except for trying to apply rules for absolutely convergent series to your series, which yields a known result that has meaning in some contexts." – 2017-02-12
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    The absolute convergence or lack thereof does not come into play here. The lack of uniform convergence is a potential issue – 2017-02-12
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    @Dr.MV oops, yes... That's my bad. I meant uniform convergence. – 2017-02-12
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    @Dr.MV I addressed this point in my post. I meant it all along, but I suppose it was too implicit. Nevertheless, I must praise the OP for their work in finding the Cesaro Sum of this series in such an interesting way. – 2017-02-12