# Integral of e^-mod(x^n)

1. Oct 29, 2009

### KoopaCooper

OK, this is going back to a problem I studied back in Calculus class; it's been a few years since I graduated, so bear with me please... :)

It's easy to integrate $$\int^{\infty}_{-\infty}\\e^{-mod(x)}\\dx$$ (giving a value of 2), and with a bit of grunt-work, the same can be done for $$\int^{\infty}_{-\infty}\\e^{-mod(x^2)}\\dx$$ (giving the value of $$\sqrt{\pi}$$). But I was curious about $$\int\\e^{-mod(x^n)}$$ for values of n>2.
Would it be considered valid to take the power series of the function and integrate it, then evaulate that at the end points?

For example, for the general case of x to an nth power... (oh heck...I tried to used the forum code but it got so tangled up...here's one I MSPainted earlier instead).

My question is, is this integration and evaluation valid? Have I integrated it properly? Does it count? (I never did check if the integrated power series converges). It certainly appears to work when I evaluated the partial sums of the series with x tending towards infinity, for the cases of m=1 and m=2, but is it legal?

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2. Oct 30, 2009

### mikeph

It would work for a finite sum, and I think it works for an infinite sum if the series converges (which it does for your conditions on n)... but I cannot give you an absolute answer, only that I think you are safe here.

You've switched m and n around from your introduction in the attachment!

3. Oct 30, 2009

### KoopaCooper

Ah, it's good to know I haven't completely lost my touch at calculus then. :)

I did switch them? *looks* O bother! That's what I get for using a picture that I created quite some time ago...:tongue2:

4. Oct 30, 2009

### Gib Z

Convergence is not a sufficient condition to interchange an integral and a summation: You must have uniform convergence.

I don't see how the final series can have a value? It doesn't converge! If the upper limit was some fixed value rather than infinity we could see from the alternating series test that the series converges because the terms decrease monotonically towards zero (after a certain term, at least) .

I don't understand what the "mod(x^2)" means? Could you clarify what it means, and how you got the value of 2 for when m=1 ?

It so happens that $$\int_0^{\infty} e^{-x^n} dx = \Gamma \left(1+\frac{1}{n}\right)$$.

5. Oct 30, 2009

### KoopaCooper

Huh, that was meant to be $$-mod(x)^2$$ - I'm not very good at writing neatly in this BulletinBoard maths code.

OK, I'll do this for m=1.

$$\int^{\infty}_{-\infty}\\e^{-mod(x)} = 2\int^{\infty}_{0}\\e^{-x}$$

$$= 2\int^{\infty}_{0}\sum^{\infty}_{n=0}\\\frac{-x^n}{n!}$$

$$= 2[\sum^{\infty}_{n=1}\frac{(-1)^n}{n+1}\frac{x^{n+1}}{n!}]\\\|^{\infty}_{x=0}$$

$$= 2[\sum^{\infty}_{n=1}\frac{(-1)^n\;x^{n+1}}{(n+1)!}\\\|^{\infty}_{x=0}$$

$$= -2\\e^{-x}\\\|^{\infty}_{x=0}$$

$$= -2\\e^{-\infty} - -2\\e^{-0}$$

$$= 0 + 2\1 = 2$$

But for m=2, that is, for the integral of $$e^{-mod(x)^2}$$, you can't convert the integrated power series back into a simple function, but evaluating that sum for any sufficiently large x (as x --> infinity) does apparently produce $$\sqrt{\pi}$$ (I ran it through on Excel once).

I didn't know that it could be expressed in terms of the Gamma function, that's interesting to know (my professor never mentioned that).

6. Oct 30, 2009

### Pere Callahan

So, mod mean modulus here?

$$mod(x)=|x|$$
?

7. Oct 30, 2009

### KoopaCooper

So that's how to write it! Heh, I just had a brain freeze about how to write the modulus symbol in forum code, I can't believe that.

Yes, that is what I meant, $$|x|$$.

*Slaps "n00b" sign on forehead* :p

8. Oct 30, 2009

### Pere Callahan

this implies at once

$$\int_{\mathbb{R}}{dx\,e^{-|x^n|}}=2\Gamma\left(1+\frac{1}{n}\right).$$

9. Oct 30, 2009

### KoopaCooper

Huh. For some reason, my class was never taught that identity, and our professor said that

$$\int_{\mathbb{R}}{e^{-|x^n|}dx}$$

was not solvable at all for $$n\geq3$$. Hence my trying to construct another way. O well, waste of my time.

10. Oct 30, 2009

### Gib Z

Probably meant not solvable in terms of elementary constants and the such. For integer and half integer arguments the Gamma function yields a "nice" value but no such expression has been found for other values.

If you want to get ahead or impress your teacher, try proving the relation.

11. Oct 30, 2009

### KoopaCooper

Heh, I couldn't impress him anyway. I graduated years ago and haven't kept in contact with anyone from uni, faculty or fellow students.

I've also been working on/off on trying to find the general derivative of the function $$y = \\^{n}\\x$$ (tetration, positive integer n only), and it seems to be defined recursively and not in a nice easy way either (involving more than one summation sign).
Although differentiating $$y = \\^{\infty}\\x$$ is much easier, as it essentially boils down to differentiating the equation $$y=x^y$$. It's a slightly messy answer:
$$y' = \frac{y^2}{x-x\ln(y)}$$
if I recall correctly, but by far easier than for finite power towers of x.