Is the Integral of an Analytic Function Divergent at Infinity?

[eljose]

let be the analytic everywhere function f(x) with limit tending to +oo and -oo with oo0 infinite then we want to calculate the integral..

$$\int_{0}^{\infty}dxe^{-x^{2}}=0.5\sqrt{\pi}$$

ot do so we expand the exponential function into a power series (we can do it as the function is analytic everywhere) so we have...

$$exp(-x^2)=\sum_{n=0}^{\infty}a_{n}x^{n}$$

but the integral of this power series is divergent in the form:

$$\sum_{n=0}^{\infty}a_{n}(\infty)^{n}$$

wich is clearly infinite...so where is the solution to this paradox?..thanks.

 

[benorin]

$$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$
expand the exponential function into a power series (we can do it as the function is analytic everywhere) so we have...
$$exp(-x^2)=\sum_{n=0}^{\infty}a_{n}x^{n}$$

.
hold-it, try:
.
$$exp(-x^2)=\sum_{n=0}^{\infty}\frac{(-1)^{n}x^{2n}}{n!}$$
.

but the integral of this power series is divergent in the form:
$$\sum_{n=0}^{\infty}a_{n}(\infty)^{n}$$
wich is clearly infinite...so where is the solution to this paradox?..thanks.

.
It's no longer clearly divergent, it converges to 0 since $$exp(-x^2)=\frac{1}{exp(x^2)}$$ goes to zero as x -> $$\infty$$.

 

[matt grime]

i like the way you start by setting f to be some every analytic function and then never mention f again after defining it. putting x=infinity into something arbitrary with no other information is rarely a good idea for determining anything.

 

[eljose]

Yes but you have the series:

$$\sum_{n=0}^{\infty}\frac{(-1)^{n}{\infty}^{2n}}{n!}$$

that you can check that is infinite...

 

[matt grime]

That series is completely meaningless, what with infinity not being a real (or complex) number and everything.

 

[shmoe]

Yes but you have the series:
$$\sum_{n=0}^{\infty}\frac{(-1)^{n}{\infty}^{2n}}{n!}$$
that you can check that is infinite...

You're trying to treat infinity like a number here and substitute it into a power series. This is not valid, it's nonsense.

 

[Zurtex]

Yes but you have the series:
$$\sum_{n=0}^{\infty}\frac{(-1)^{n}{\infty}^{2n}}{n!}$$
that you can check that is infinite...

Do you understand what a limit is?

 

[benorin]

Here it is. Gamma(1/2).

To prove that $$\int_{0}^{\infty}e^{-x^{2}}dx=\frac{\sqrt{\pi}}{2}$$
Put $$I=\int_{0}^{\infty}e^{-x^{2}}dx$$ so that $$I^2=\int_{x=0}^{\infty}e^{-x^{2}}dx\int_{y=0}^{\infty}e^{-y^{2}}dy=\int_{y=0}^{\infty}\int_{x=0}^{\infty}e^{-(x^{2}+y^{2})}dxdy$$.
Now Transform to polar coordinates, and note that the first quadrant (e.g. QI) is one quarter of an infinite plane in rectangluar coordinates, so too is it one quarter of an infinite circle in polar coordinates (you can prove it with using squeeze theorem if your so inclined); you get
$$I^2=\int_{\theta=0}^{\frac{\pi}{2}}\int_{r=0}^{\infty}e^{-r^{2}}rdrd\theta=\frac{1}{2}\int_{\theta=0}^{\frac{\pi}{2}}d\theta\int_{u=0}^{\infty}e^{-u}du=\frac{\pi}{4}\lim_{M \rightarrow \infty}(1-e^{-M})=\frac{\pi}{4}$$
Therefore, $$I=\frac{\sqrt{\pi}}{2}$$.
To explain the title of my post: the Euler Gamma function can be defined as follows: $$\Gamma(z)=\int_{t=0}^{\infty}e^{-t}t^{z-1}dt$$ which converges
$$\forall z\in \mathhbb{C}$$ such that $$\Re{z}>0$$. The given integral is equal to $$\Gamma(\frac{1}{2})$$.
The Gamma function is the well studied analytic continuation of the factorial to all complex values except non-positive integers. In particular, $$\Gamma(n)=(n-1)!, \forall n\in\mathbb{N}$$.
Enjoy.

 

[eljose]

Yes but let,s suppose we have the integral:
$$\int_{0}^{\infty}dxe^{-{cos(x)}}$$
you can see that exp(-cos(x)) is analytic everywhere so we can substitute the function by its Taylor series in the form:
$$exp(-cos(x))=\sum_{n=0}^{\infty}a(n)x^{n}$$ there is no error in that so we would have that the integral would be the limit:
$$Lim k--->\infty\sum_{n=0}^{\infty}a(n)k^{n}$$

 

[matt grime]

What are you trying to do? If it were to abuse latex you're succeeding. \lim will produce limits proper, and \text{foo} produces foo as if it were text. Now, how about trying to explain what it is you want to know? Are you just trolling now? OUt of curiosity (and I can never remember the answer) under what conditions are we allowed to interchange function with its taylor series, the integral of a sum with the sum of the integrals, and the limit of a sum with the sum of the limits?

 

[masudr]

1. You can exchange a function with it's taylor series whenever the function is analytic in that region (sometimes that's how we define analytic!).

2. The integral of a sum is always the of the integrals.

3. The limit of a sum is the sum of the limits if... I can't remember this one.

 

[benorin]

Yes but let,s suppose we have the integral:
$$\int_{0}^{\infty}dxe^{-{cos(x)}}$$
you can see that exp(-cos(x)) is analytic everywhere so we can substitute the function by its Taylor series in the form:
$$exp(-cos(x))=\sum_{n=0}^{\infty}a(n)x^{n}$$ there is no error in that so we would have that the integral would be the limit:
$$\lim_{k\rightarrow\infty}\sum_{n=0}^{\infty}b(n)k^{n}$$

Yes, that is because this integral is divergent. I changed you sum to have b(n) as some integration is necessary, not just taking the limit of the power series, and, by the way, uniform convergence is required for the interchange of limit operations (e.g., to say that the integral of an infinite sum is the infinite sum of the integrals requires uniform convergence of the power series, not analyticity).

But you have something of a valid concern, perhaps this would help you:

$$e^{-x^{2}}=\sum_{n=0}^{\infty}\frac{(-1)^{n}{x}^{2n}}{n!}$$

does indeed converge on every bounded closed interval, say [-M,M], per the ratio test as

$$\lim_{n\rightarrow\infty}|\frac{(-1)^{n+1}{x}^{2n+2}}{(n+1)!}\cdot\frac{n!}{(-1)^{n}{x}^{2n}}|=\lim_{n\rightarrow\infty}\frac{{x}^{2}}{n+1}=0,\forall x\in(-\infty,\infty)$$ this limit is zero for all finite x, and hence the series converges on every bounded closed interval.
It is also of note that the above series for $$e^{-x^{2}}$$ converges uniformly on every bounded closed interval, (but not on unbounded ones that admit infinities). This accounts for the following facts:

While $$\lim_{x\rightarrow\infty}\lim_{N\rightarrow\infty}\sum_{n=0}^{N}\frac{(-1)^{n}{x}^{2n}}{n!}=0$$, suprisingly
$$\lim_{N\rightarrow\infty}\lim_{x\rightarrow\infty}\sum_{n=0}^{N}\frac{(-1)^{n}{x}^{2n}}{n!}=-\infty$$; this is because the convergence of the series is NOT uniform on unbounded intervals (when x is an infinity).