Math Help - Another nasty integral

1. Originally Posted by CaptainBlack
Rough outline:

Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$

1. Every function can be written as the sum of an odd function and an even function.

2. The result is trivially true for a sufficiently well behaved odd function.
Both sides are zero for a sufficiently well behaved odd function.

3. The result is a result of the fundamental theorem of calculus, and the
Laplace transform of a derivative for an even function zero at the origin(sufficiently well behaved).
For an even function $f(x)$ such that $f(0)=0$

$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)+f(0)=$ $2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)$

.............. $=2\int_{0}^{\infty}f'(x) dx=2\left[\lim_{x \to \infty}f(x) - \lim_{x \to 0+}f(x)\right]$

but as $f$ is even (and sufficiently well behaved) the last limit
above is $0$, and $\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)$

For a constant we have:

$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx\ k\ e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx\ k\ e^{-\epsilon |x|}=$ $2 \lim_{\epsilon \to 0+} \epsilon\ \mathcal{L}k(\epsilon)=2 \lim_{\epsilon \to 0^+} \epsilon\ \frac{k}{\epsilon}=2\ k
$
,

which is the required result in this case.

(Note we could have done this for an arbitrary even function by keeping track of the f(0) terms rather than setting them to zero, then we would not have had to handle the constant term separately)

4. Hence the result is true for a sufficiently well behaved function.

I can amplify the details if you need (though I doubt that I will feel like
going for full rigour)

RonL
RonL

2. Originally Posted by CaptainBlack
Rough outline:

1. Every function can be written as the sum of an odd function and an even function.
Obvious but here goes, for any function $f(x)$ on $\mathbb{R}$ define the even and odd functions $g(x)$ and $h(x)$ as:

$g(x)=\left[f(x)+f(-x)\right]/2$

$h(x)=\left[f(x)-f(-x)\right]/2$,

then:

$f(x)=g(x)+h(x)$

I hope this is all very familiar from other contexts

RonL

3. Originally Posted by CaptainBlack
Rough outline:

Given a sufficiently smooth function f(x) we have that:
$\lim_{x \to \infty}f(x) + \lim_{x \to - \infty}f(x) = \lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}$
We need that if (sufficiently well behaved) functions $f_1(x)$ and $f_2(x)$ satisfy the above relation then so does any linear combination of them.

This is obvious and I will not prove it, but it needs stating for the demonstration to go through when we split a function into its odd and even components.

RonL

4. Originally Posted by CaptainBlack
Both sides are zero for a sufficiently well behaved odd function.

For an even function $f(x)$

$\lim_{\epsilon \to 0^+} \epsilon \int_{-\infty}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \epsilon \int_{0}^{\infty} dx f(x) e^{-\epsilon |x|}=2 \lim_{\epsilon \to 0^+} \left[ \mathcal{L}f' \right] (\epsilon)$

.............. $=2\int_{0}^{\infty}f'(x) dx=2\left[\lim_{x \to \infty}f(x) - \lim_{x \to 0+}f(x)\right]$

but as $f$ is even (and sufficiently well behaved) the last limit
above is $0$, and $\lim_{x \to \infty}f(x)=\lim_{x \to -\infty}f(x)$
There is a slight error here, in that I have assumed that the even function is zero at $x=0$, which is not necessarily true.

Instead supose we decompose the original function into the sum of a constant, and odd function and an even function which is zero at the origin (an alternative is to leave the decomposition into odd and even, and kept track of the f(0) terms in the even function analysis when they would have cancelled out).

The relation is satisfied by the constant, the odd function and the even function zero at the origin, so the result holds.

I have gone back and modified the previous posts to reflect this change.

RonL

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