integral of |x|^-|x| from -infinity to infinity

[rakuenso]

^ as title states

 

anyone?

[Dave]

Are you sure this can even be evaluated? I mean, one approach would certainly be to use the fact that:

 

[math]\int_{-\infty}^{\infty} |x|^{|x|} \, dx = \int_{-\infty}^{0} (-x)^{x} \, dx = + \int_{0}^{\infty} x^{(-x)} \, dx[/math]

 

Besides this, I don't think that function is integrable at all (although I could very well be wrong). Is this an assignment, or is this just something to satisfy curiosity?

[Severian]

Use symmetry first:

 

[math] \int_{-a}^{a} |x| e^{-|x|} dx = 2 \int_0^a x e^{-x} dx [/math]

 

Now [math] \int_0^a x e^{-x} dx = \left[- e^{-x} (1+x) \right]_0^a = 1- (1+a)e^{-a}[/math] (integration by parts to see this)

 

So [math]\int_{-a}^{a} |x| e^{-|x|} dx = 2 \left( 1- (1+a)e^{-a} \right)[/math]

 

Finally, taking [math] a \to \infty[/math],

 

[math] \int_{-\infty}^{\infty} |x| e^{-|x|} dx = \lim_{a \to \infty} 2 \left( 1- (1+a)e^{-a} \right) = 2[/math]

[Dave]

Where's the exp come from?

[Severian]

oops, I misread the question. My bad

[Tom Mattson]

[math]\int_{-\infty}^{\infty} |x|^{|x|} \' date=' dx = \int_{-\infty}^{0} (-x)^{x} \, dx = + \int_{0}^{\infty} x^{(-x)} \, dx[/math']

 

I don't see why there is an equals sign between the last two integrals. Surely if you are cutting the region of integration at zero, you must add the two integrals, right?

 

I would think that it should be either this:

 

[math]\int_{-\infty}^{\infty} |x|^{|x|} \, dx = \int_{-\infty}^{0} (-x)^{x} \, dx + \int_{0}^{\infty} x^{(-x)} \, dx[/math]

 

or this:

 

[math]\int_{-\infty}^{\infty} |x|^{|x|} \, dx = 2\int_{-\infty}^{0} (-x)^{x} \, dx = 2 \int_{0}^{\infty} x^{(-x)} \, dx[/math]

[rakuenso]

note that is |x|^NEGATIVE|x|, its to satisfy curiosity.

[Tom Mattson]

note that is |x|^NEGATIVE|x|' date='

[/quote']

 

We know that, and that is exactly how we treated it.

 

Don't forget that:

 

-|x|=x when x<0, and

-|x|=-x when x>0.

 

Edit:

 

Ah, I see what you mean. You are referring to the fact that dave wrote the original integral incorrectly. That was just a typo (which I didn't even notice as I copied and pasted it twice!). But even after that typo is fixed, nothing else changes.

 

So, just for completeness and accuracy here are my last two equations, corrected:

 

[math]

\int_{-\infty}^{\infty} |x|^{-|x|} \, dx = \int_{-\infty}^{0} (-x)^{x} \, dx + \int_{0}^{\infty} x^{(-x)} \, dx

[/math]

 

[math]

\int_{-\infty}^{\infty} |x|^{-|x|} \, dx = 2\int_{-\infty}^{0} (-x)^{x} \, dx = 2 \int_{0}^{\infty} x^{(-x)} \, dx

[/math]

[Dave]

I don't see why there is an equals sign between the last two integrals. Surely if you are cutting the region of integration at zero, you must add the two integrals, right?

 

You're right, of course I was copying and pasting, so must have pasted the = by accident