Solving Difficult Integral: Exploiting Orthogonality of Plane Waves

[Morberticus]

I'm trying to sort out the integral:

\int x^{-2}e^{ikx}dx.

At first, I thought I would have to solve it numerically. But I am hoping to exploit the orthogonality of the plane waves:

\int e^{ikx}dx = 2\pi\delta(k)

It has been a while since I have dealt with integral manipulation, so any help would be appreciated.

 

[Dickfore]

what are the limits?

 

[Morberticus]

what are the limits?

The limits are at minus infinity and infinity.

[edit]-actually good point. If the integral is infinite then I will have to introduce limits.

 

[Dickfore]

The integral is not convergent in the Riemann sense. It has a pole of order 2 at x = 0. Nevertheless, you can formally try this trick. Consider the integral as a function of the parameter k:

<br /> f(k) \equiv \int_{-\infty}^{\infty}{x^{-2} \, e^{i \, k \, x} \, dx}<br />

Take two derivatives with respect to k. What do you get?

 

[Morberticus]

The integral is not convergent in the Riemann sense. It has a pole of order 2 at x = 0. Nevertheless, you can formally try this trick. Consider the integral as a function of the parameter k:

<br /> f(k) \equiv \int_{-\infty}^{\infty}{x^{-2} \, e^{i \, k \, x} \, dx}<br />

Take two derivatives with respect to k. What do you get?

Hmmm... My guess would be

f&#039;&#039;(k) = \int^{\infty}_{-\infty} e^{ikx}dx

which is infinite, right?

It's funny that you mention f(k) because that is precisely what I would need, as the integral will be nested in a larger, numerical integral over k, and I was hoping to save on a lot of computational time by getting f(k) analytically.

 

[Dickfore]

You made a sign error and the integral you got is of the same form as one of the integrals you had already posted.

 

[Morberticus]

You made a sign error and the integral you got is of the same form as one of the integrals you had already posted.

Ah yes, I forgot to bring down the 'i' twice (I'm horribly rusty).

So I would get:

f&#039;&#039;(k) = -\int^{\infty}_{-\infty}e^{ikx}dx = -2\pi\delta(k)

Though is there a way I can implement this? Would I be allowed to, say, do the following:

f(k) = \frac{2\pi k^2}{2}\delta(k)

I.e. re-integrate f''(k)

 

[Dickfore]

Yes, you are allowed to re-integrate, but your integration is incorrect.

 

[Morberticus]

Yes, you are allowed to re-integrate, but your integration is incorrect.

Forgot completely that delta was a function of k.

So would these steps be correct?

f&#039;(k) = 2\pi

f(k) = 2\pi k

[edit]-Oops, sign error. But is the form correct?

 

[Dickfore]

Hint:

Consider the Heaviside step function:

<br /> \theta (k) = \left\{\begin{array}{rcl}<br /> 1 &amp;,&amp; k \ge 0 \\<br /> <br /> 0 &amp;,&amp; k &lt; 0<br /> \end{array}\right.<br />

What is the derivative of this function?

 

[Morberticus]

Hint:

Consider the Heaviside step function:

<br /> \theta (k) = \left\{\begin{array}{rcl}<br /> 1 &amp;,&amp; k \ge 0 \\<br /> <br /> 0 &amp;,&amp; k &lt; 0<br /> \end{array}\right.<br />

What is the derivative of this function?

Hmmm... The derivative is a delta function. But would that mean I would have to manipulate something like:

\theta(\infty) = \int^{\infty}_{-\infty} \delta(k)dk

as

f&#039;&#039;(k) = -2\pi\delta(k)

f&#039;(k) = -2\pi \int^{\infty}_{-\infty} \delta(k)dk = -2\pi\theta(\infty)

[edit]- Have to run but will be back tomorrow. Thanks a million. This should save me weeks of cpu time.

 

[Mute]

Hmmm... The derivative is a delta function. But would that mean I would have to manipulate something like:

\theta(\infty) = \int^{\infty}_{-\infty} \delta(k)dk

as

f&#039;&#039;(k) = -2\pi\delta(k)

f&#039;(k) = -2\pi \int^{\infty}_{-\infty} \delta(k)dk = -2\pi\theta(\infty)

[edit]- Have to run but will be back tomorrow. Thanks a million. This should save me weeks of cpu time.

When you integrate f''(k) you want a function of k again, which you don't get if you integrate over the entire k range. If you integrate over the entire range of k there's no k left.

f&#039;(k) = f&#039;(-\infty) - 2\pi \int_{-\infty}^k dk&#039;~\delta(k&#039;) = -2\pi \theta(k)
assuming f and f' vanish at the boundaries. You then do something similar for the next integration.

 

[Morberticus]

When you integrate f''(k) you want a function of k again, which you don't get if you integrate over the entire k range. If you integrate over the entire range of k there's no k left.

f&#039;(k) = f&#039;(-\infty) - 2\pi \int_{-\infty}^k dk&#039;~\delta(k&#039;) = -2\pi \theta(k)
assuming f and f' vanish at the boundaries. You then do something similar for the next integration.

I have read up a little on the method you have used (dummy variable?) and I think I have the hang of it. But would it be valid to assume that f(k) and f'(k) are bounded? Would this need to be tested?

The integral has shown up due to the Fourier transform:

|r-r&#039;|^{-1} = (1/2\pi^2)\int x^{-2} e^{ix(r-r&#039;)}dx

My intuition is telling me that, as (r-r') gets larger, the integral should get smaller, as |r-r'|^-1 will get smaller. Or, to use the variable I have been using up to now, |k|^-1 goes to 0 as k goes to infinity. Would this also imply that f(k) goes to infinity as k goes to 0? If it does then I am an idiot, as I am integrating an expression that looks like:

\frac{1}{2\pi^2}\int^{a}_{b}sin(Cr)sin(Dr&#039;)[\int x^{-2} e^{ix(r-r&#039;)}dx]sin(Er)sinF(r&#039;)drdr&#039;and the purpose of the nested integral was ultimately to avoid an infinity when r = r'. I am just copping that the function f(k) might be infinite at k = 0? In that case, I am assuming adding finite limits to the nested integral would give me a finite approximation of f(k) at k = 0 which would suit my purposes fine.

Thanks to both you and Dickfore for your help so far, as I think I better understand how to manipulate the integral. My (hopefully) last question is, is the same line of logic you have shown me valid if the limits are finite?

[edit]- Actually, would I be right in thinking it would be easier, as all I would need is an expression as a function of k and two arbitrary bounds?

 

[Morberticus]

I have tried integrating the intral in mathematica, and it has produced the expression:

\int x^{-2} e^{ikx}dx = ikEI(ikx) - \frac{e^{ikx}}{x}

Ei(ikx) has a simple enough expression that I can approximate with a series expansion. Here is hoping that it will then simply be a case of inserting the two arbitrary bounds in place of x.

 

[Dickfore]

I have tried integrating the intral in mathematica, and it has produced the expression:

\int x^{-2} e^{ikx}dx = ikEI(ikx) - \frac{e^{ikx}}{x}

Ei(ikx) has a simple enough expression that I can approximate with a series expansion. Here is hoping that it will then simply be a case of inserting the two arbitrary bounds in place of x.

What limits did you use though?

 

[Dickfore]

When you integrate f''(k) you want a function of k again, which you don't get if you integrate over the entire k range. If you integrate over the entire range of k there's no k left.

f&#039;(k) = f&#039;(-\infty) - 2\pi \int_{-\infty}^k dk&#039;~\delta(k&#039;) = -2\pi \theta(k)
assuming f and f' vanish at the boundaries. You then do something similar for the next integration.

For the last integration:

<br /> f(k) = f(-\infty) + \int_{-\infty}^{k}{f&#039;(t) \, dt}<br />

<br /> f(k) = -2 \, \pi \, \int_{-\infty}^{k}{\theta(t) \, dt}<br />

I would use integration by parts.

 

[Morberticus]

What limits did you use though?

I assumed the limits could be inserted the usual way, to get the expression:

f(k) = ik(Ei(ika)-Ei(ikb)) + (\frac{e^{ikb}}{b}-\frac{e^{ika}}{a})

Where a and b are arbitrary limits.

 

[Morberticus]

Hmmm... I was hoping that, as the interval |a-b| approaches infinity, the integral would approach |k|^-1

But when I plot f(k) from my previous post, the real component oscillates, and the imaginary component appears to be quadratic in form. Increasing |a-b| only increases the frequency of oscillation.

I will give the integration by parts a try.

 

[Morberticus]

For the last integration:

<br /> f(k) = f(-\infty) + \int_{-\infty}^{k}{f&#039;(t) \, dt}<br />

<br /> f(k) = -2 \, \pi \, \int_{-\infty}^{k}{\theta(t) \, dt}<br />

I would use integration by parts.

They way I've approached it:

\int^{\infty}_{-\infty}\theta(t)dt = \int^{0}_{-\infty}\theta(t)dt + \int^{k}_{0}\theta(t)dt

And since \theta(k) = 1 for all k > 0:

\int^{0}_{-\infty}\theta(t)dt + \int^{k}_{0}\theta(t)dt = 0 + \int^{k}_{0}1dt = k

There is a fudge: I have divided the integral (-inf,inf) into (-inf,0] (0,inf] so 0 technically isn't a member of the second interval.

 

[Dickfore]

Why do you take the upper limit of the integral to be \infty?

 

[Morberticus]

Why do you take the upper limit of the integral to be \infty?

Oops, typo. The limits are -infinity and k, and I divide the integral into -infinity,0 and 0,k.

 

[Dickfore]

What happens if k &lt; 0?

 

[Morberticus]

What happens if k &lt; 0?

Hmm... If k < 0 then I am guessing the integral is 0, as \theta(t) = 0 across the interval.

So would I end up with:

f(k) = -2\pi\int^{k}_{-\infty}\theta(t)dt = -2\pi k\theta(k)

 

[italian1013]

doesnt the person only have to do intergration by parts:

UV - ∫vdu

to solve the problem?


or am i thinking of something else?

 

[Petar Mali]

Just to say

\int^{\infty}_{-\infty}e^{ikx}dx diverge.

 

[Dickfore]

Hmm... If k < 0 then I am guessing the integral is 0, as \theta(t) = 0 across the interval.

So would I end up with:

f(k) = -2\pi\int^{k}_{-\infty}\theta(t)dt = -2\pi k\theta(k)

Hmm, if you take the inverse Fourier transform of this:

<br /> \int_{-\infty}^{\infty}{(-2 \, \pi \, k \, \theta(k) \, e^{i k x} \, \frac{d k}{2 \pi}}<br />

I don't think you will get x^{-2}.

The problem with the convergence properties of the integral is, of course, the pole of order 2 on the real axis at x = 0. Let's try something different: Let's consider x as a complex variable and shift the pole either above or below the integration path by an infinitesimal amount and let us close the contour of integration either along a large semicircle in the upper or lower part of the imaginary plane. To see where to close it, consider the absolute value of the factor:

<br /> |\exp(-i \, k \, z)| = e^{k \, \Im{(z)}}<br />

and require that it tends to zero when z \rightarrow \infty. We see that we must have k \, \Im{(z)} &lt; 0, i.e. close it in the upper half-plane if k &lt; 0 (notice that your contour is in the counterclockwise direction) or lower half-plane if k &gt; 0 (the contour is in the clockwise direction). Also, the residue of the function:

<br /> f(z) = \frac{\exp{(-i \, k \, z)}}{(z \mp i \, \epsilon)^{2}}<br />

at the residue at the double pole is:

<br /> \mathrm{Res}\left[f(z), z = \pm i \, \epsilon \right] = \lim_{z \rightarrow \pm i \, \epsilon}{\frac{d}{d z}\left(\exp{(-i \, k \, z)} \right)} = -i \, k \, e^{\pm k \, \epsilon}<br />

Make sure you shift the pole so that it is inside the contour, i.e. take +i \, \epsilon if k&lt; 0 or -i \, \epsilon if k&gt;0. Both of these cases can be written as -i \, \mathrm{sgn}(k) \, \epsilon.

Noticing that the integral has a zero contribution over an infinite semi-circle, we have:

<br /> \int_{-\infty}^{\infty}{(x + i \, \mathrm{sgn}(k) \, \epsilon)^{-2} \, e^{-i \, k \, x} \, dx} = \oint_{C_{k}}{\frac{\exp(-i \, k \, z)}{(z + i \, \mathrm{sgn}(k) \, \epsilon)^{2}} \, dz} = -\mathrm{sgn}(k) \, 2 \pi \, i \, \mathrm{Res}\left[f(z), z = -\mathrm{sgn}(k) \, i \, \epsilon \right] = -2 \, \pi \, k \, \mathrm{sgn}(k) \, e^{-\mathrm{sgn}(k) \, k \, \epsilon} = -2 \, \pi \, |k| \, e^{-|k| \, \epsilon}<br />

Finally, taking the limit \epsilon \rightarrow 0+, we get:

<br /> \int_{-\infty}^{\infty}{x^{-2} \, e^{-i \, k \, x} \, dx} = -2 \pi \, |k|<br />

When I perform the inverse Fourier transform, I get 2/x^{2}, so, maybe you need to take half this value.

 

[Morberticus]

Hmm, if you take the inverse Fourier transform of this:

<br /> \int_{-\infty}^{\infty}{(-2 \, \pi \, k \, \theta(k) \, e^{i k x} \, \frac{d k}{2 \pi}}<br />

I don't think you will get x^{-2}.

The problem with the convergence properties of the integral is, of course, the pole of order 2 on the real axis at x = 0. Let's try something different: Let's consider x as a complex variable and shift the pole either above or below the integration path by an infinitesimal amount and let us close the contour of integration either along a large semicircle in the upper or lower part of the imaginary plane. To see where to close it, consider the absolute value of the factor:

<br /> |\exp(-i \, k \, z)| = e^{k \, \Im{(z)}}<br />

and require that it tends to zero when z \rightarrow \infty. We see that we must have k \, \Im{(z)} &lt; 0, i.e. close it in the upper half-plane if k &lt; 0 (notice that your contour is in the counterclockwise direction) or lower half-plane if k &gt; 0 (the contour is in the clockwise direction). Also, the residue of the function:

<br /> f(z) = \frac{\exp{(-i \, k \, z)}}{(z \mp i \, \epsilon)^{2}}<br />

at the residue at the double pole is:

<br /> \mathrm{Res}\left[f(z), z = \pm i \, \epsilon \right] = \lim_{z \rightarrow \pm i \, \epsilon}{\frac{d}{d z}\left(\exp{(-i \, k \, z)} \right)} = -i \, k \, e^{\pm k \, \epsilon}<br />

Make sure you shift the pole so that it is inside the contour, i.e. take +i \, \epsilon if k&lt; 0 or -i \, \epsilon if k&gt;0. Both of these cases can be written as -i \, \mathrm{sgn}(k) \, \epsilon.

Noticing that the integral has a zero contribution over an infinite semi-circle, we have:

<br /> \int_{-\infty}^{\infty}{(x + i \, \mathrm{sgn}(k) \, \epsilon)^{-2} \, e^{-i \, k \, x} \, dx} = \oint_{C_{k}}{\frac{\exp(-i \, k \, z)}{(z + i \, \mathrm{sgn}(k) \, \epsilon)^{2}} \, dz} = -\mathrm{sgn}(k) \, 2 \pi \, i \, \mathrm{Res}\left[f(z), z = -\mathrm{sgn}(k) \, i \, \epsilon \right] = -2 \, \pi \, k \, \mathrm{sgn}(k) \, e^{-\mathrm{sgn}(k) \, k \, \epsilon} = -2 \, \pi \, |k| \, e^{-|k| \, \epsilon}<br />

Finally, taking the limit \epsilon \rightarrow 0+, we get:

<br /> \int_{-\infty}^{\infty}{x^{-2} \, e^{-i \, k \, x} \, dx} = -2 \pi \, |k|<br />

When I perform the inverse Fourier transform, I get 2/x^{2}, so, maybe you need to take half this value.

Strangely enough, I have been messing around with mathematica and am getting something vaguely similar. Thanks.