How Does the Sinc Function Integral Relate to Quantum Collision Theory?

• Santilopez10
In summary, the problem is to show that the given integral equals 0 for |p| ≥ 1 and π for |p| < 1, and to discuss the cases when p = ±1. The solution involves considering the integral in the complex plane and using line integrals. The smaller semicircle is parametrized and the residue theorem is applied. The upper bound for the larger semicircle is estimated using the half circle parametrization and absolute values. However, the difficulty lies in changing variables in the real axis integrals back to the original integral.
Santilopez10

Homework Statement

The following is a problem from "Applied Complex Variables for Scientists and Engineers"
It states:
The following integral occurs in the quantum theory of collisions:
$$I=\int_{-\infty}^{\infty} \frac {sin(t)} {t}e^{ipt} \, dt$$
where p is real. Show that
$$I=\begin{cases}0 & \text{if } |p| \geq 1 \\\pi & \text{if } |p| < 1\end{cases}$$
Discuss the cases when ##p= \pm 1##

Homework Equations

ML inequality and Cauchy theorem

The Attempt at a Solution

First, I consider this integral in the complex plane:
$$\oint_C \frac {e^{iz}} {z} e^{ipz} \, dz$$
where C is the contour defined as a semicircular path of radius R with a cut at the origin, as a semicircular path of radius ##\epsilon##.
Okey, first, we see that no singularity is within the contour, so the integral as a whole is just 0. Then, I proceed to write the integral as different line integrals.
$$\oint_C \frac {e^{iz}} {z} e^{ipz} \, dz = \int_{\Gamma}\ + \int_{-R}^{\epsilon} \frac {e^{it}} {t} e^{ipt} \, dt + \int_{\gamma}\ + \int_{\epsilon}^{R} \frac {e^{it}} {t} e^{ipt} \, dt = 0$$
Where ##\Gamma## is the larger semicircle, and ##\gamma## is the smaller. I expressed the integrals with variable ##t## because they lie in the real axis.

I start with the smaller semicircle, it can be parametrized as ##z=\epsilon e^{i\phi}, dz=i \epsilon e^{i\phi}## and it goes from ##\phi = \pi## to ##\phi = 0##
There is a theorem that states that the value of the integral is just ##-i\pi Residue(f(z))## and in this case the residue its just 1, so that integral is done.
Now here is where I am stuck with 2 problems:

1) How do I change variables in the Real axes integrals so that I can get back the original integral?
2) This is just trying to know if my estimation of the upper bound of the large semicircle is correct:
I begin using the common half circle parametrization so that I end up with:
$$\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta$$
and so I use absolute values to find an upper bound for the integral:
$$|\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta|=\int_{0}^{\pi} |i e^{i(p+1)R e^{i\theta}}|\, d\theta$$
then:
$$\int_{0}^{\pi} |i| |e^{i(p+1)R e^{i\theta}}|\, d\theta$$
and
$$\int_{0}^{\pi} |e^{i(p+1)R e^{i\theta}}|\, d\theta = \int_{0}^{\pi}|e^{i(p+1)R (cos(\theta)+isin(\theta)}|\, d\theta =\int_{0}^{\pi} |e^{i(p+1)R cos(\theta)}||e^{-R(p+1)sin(\theta)}|$$
and then
$$\int_{0}^{\pi} |e^{-R(p+1)sin(\theta)}|\, d\theta$$
then by using the fact the integral is simmetric respect to ##\frac {\pi}{2}##:
$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)sin(\theta)}|\, d\theta$$
and because ##-sin(\theta)\leq -\frac{2\theta}{\pi}, \theta \in (0,\frac{\pi}{2})## we can write the integral as:
$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)\frac{2\theta}{\pi}}|\, d\theta$$
we can drop the absolute values, and the result of the integral is:
$$\frac{\pi(1-e^{-R(p+1)})}{R(p+1)}$$
which I suppose approaches different values depending on the value of p, still need to check that (if it does not, better, it means it goes to 0 for any value of p and simplifies everything). But still, I would be grateful if someone could tell me if the estimation is correct.

As it seems, the thing that does not let me continue is the way to return to the original integral.

Delta2
Santilopez10 said:

Homework Statement

The following is a problem from "Applied Complex Variables for Scientists and Engineers"
It states:
The following integral occurs in the quantum theory of collisions:
$$I=\int_{-\infty}^{\infty} \frac {sin(t)} {t}e^{ipt} \, dt$$
where p is real. Show that
$$I=\begin{cases}0 & \text{if } |p| \geq 1 \\\pi & \text{if } |p| < 1\end{cases}$$
Discuss the cases when ##p= \pm 1##

Homework Equations

ML inequality and Cauchy theorem

The Attempt at a Solution

First, I consider this integral in the complex plane:
$$\oint_C \frac {e^{iz}} {z} e^{ipz} \, dz$$
where C is the contour defined as a semicircular path of radius R with a cut at the origin, as a semicircular path of radius ##\epsilon##.
Okey, first, we see that no singularity is within the contour, so the integral as a whole is just 0. Then, I proceed to write the integral as different line integrals.
$$\oint_C \frac {e^{iz}} {z} e^{ipz} \, dz = \int_{\Gamma}\ + \int_{-R}^{\epsilon} \frac {e^{it}} {t} e^{ipt} \, dt + \int_{\gamma}\ + \int_{\epsilon}^{R} \frac {e^{it}} {t} e^{ipt} \, dt = 0$$
Where ##\Gamma## is the larger semicircle, and ##\gamma## is the smaller. I expressed the integrals with variable ##t## because they lie in the real axis.

I start with the smaller semicircle, it can be parametrized as ##z=\epsilon e^{i\phi}, dz=i \epsilon e^{i\phi}## and it goes from ##\phi = \pi## to ##\phi = 0##
There is a theorem that states that the value of the integral is just ##-i\pi Residue(f(z))## and in this case the residue its just 1, so that integral is done.
Now here is where I am stuck with 2 problems:

1) How do I change variables in the Real axes integrals so that I can get back the original integral?
2) This is just trying to know if my estimation of the upper bound of the large semicircle is correct:
I begin using the common half circle parametrization so that I end up with:
$$\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta$$
and so I use absolute values to find an upper bound for the integral:
$$|\int_{0}^{\pi} i e^{i(p+1)R e^{i\theta}}\, d\theta|=\int_{0}^{\pi} |i e^{i(p+1)R e^{i\theta}}|\, d\theta$$
then:
$$\int_{0}^{\pi} |i| |e^{i(p+1)R e^{i\theta}}|\, d\theta$$
and
$$\int_{0}^{\pi} |e^{i(p+1)R e^{i\theta}}|\, d\theta = \int_{0}^{\pi}|e^{i(p+1)R (cos(\theta)+isin(\theta)}|\, d\theta =\int_{0}^{\pi} |e^{i(p+1)R cos(\theta)}||e^{-R(p+1)sin(\theta)}|$$
and then
$$\int_{0}^{\pi} |e^{-R(p+1)sin(\theta)}|\, d\theta$$
then by using the fact the integral is simmetric respect to ##\frac {\pi}{2}##:
$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)sin(\theta)}|\, d\theta$$
and because ##-sin(\theta)\leq -\frac{2\theta}{\pi}, \theta \in (0,\frac{\pi}{2})## we can write the integral as:
$$2\int_{0}^{\frac{\pi}{2}} |e^{-R(p+1)\frac{2\theta}{\pi}}|\, d\theta$$
we can drop the absolute values, and the result of the integral is:
$$\frac{\pi(1-e^{-R(p+1)})}{R(p+1)}$$
which I suppose approaches different values depending on the value of p, still need to check that (if it does not, better, it means it goes to 0 for any value of p and simplifies everything). But still, I would be grateful if someone could tell me if the estimation is correct.

As it seems, the thing that does not let me continue is the way to return to the original integral.

Suppose ##p \geq 0.## Write the integral as
$$\lim_{R \to \infty, \epsilon \to 0} \left( \int_{-R}^{-\epsilon} + \int_{\epsilon}^R \right) \frac{1}{x} \frac{1}{2i} \left( e^{ix + ip x} - e^{-ix + ipx} \right) \, dx$$
Note that if we keep ##\sin(x)## intact, we can integrate through ##0## because ##\sin(x)/x## has no singularity at ##x \to 0##: this function can be defined as ##1## when ##x = 0.## However, when we split up ##\sin(x)## into exponentials, we need to keep away from ##x=0##, hence the ##(-\epsilon,+\epsilon)## exclusion interval. Let ## I_1 ## be the integral containing ##e^{ix + ip x}## and ##I_2## be the other one, containing ##e^{-ix + ipx}##.

(1) Evaluation of ##I_1.## Since we are assuming ##p \geq 0##, we can complete ##I_1## in the upper ##z-##plane (that is, take ##z = x + iy, \:y \geq 0##). Define the points ##A = \epsilon +i0##, ##B = R + i0##, ##C = R + iR##, ##D = -R + iR##, ##E = -R + i0## and ##F = -\epsilon + i0##. Take the closed contour ##\gamma_1 ## equal to ##A \to B \to C \to D \to E \to F \to A##, where the last part is clockwise around the semicircle of radius ##\epsilon## in the upper half-plane. Of course, for ##f_1 = \frac{1}{2i z} e^{i(1+p)z}## we have ##\int_{\gamma_1} f_1 \, dz = 0,## as you note. Now
$$\int_B^C f_1 = \int_0^R \frac{1}{2 i (R+iy)} e^{i(1+p) (R + iy)} i \, dy,$$ so
$$\left|\int_B^C f_1 \right| \leq \int_0^R \frac{1}{2R} e^{-(1+p) y} dy \to 0 \; \text{as} \; R \to \infty.$$
Similarly, ##\int_C^D f_1 \to 0## and ##\int_D^E f_1 \to 0## as ##R \to \infty.## So, in the ##R \to \infty## limit we have ##\int_{-\infty \to F} f_1 + \int_{F \to A} f_1 + \int_{A \to \infty} f_1 = 0.## That is, ## ( \int_{-\infty \to F} + \int_{A \to \infty} ) f_1 = - \int_{F \to A} f_1,## where the last integral is clockwise around the small semi-circle. It is easy to compute in the limit ##\epsilon \to 0+.##

(2) Evaluation of ##I_2.## For the case of ##0 < p < 1## we can complete ##f_2 = \frac{1}{2iz} e^{-ix + ipx}## in the lower half-plane, much as was done above, and then compute the limiting integral. However, if ##p > 1## we must complete ##f_2## in the upper half-plane, and compute the limiting integral.

For the case of ##p = 1## we can completer ##f_1## in the upper half-plane and ##f_2## in the lower half-plane.

Delta2
Ray Vickson said:
Suppose ##p \geq 0.## Write the integral as
$$\lim_{R \to \infty, \epsilon \to 0} \left( \int_{-R}^{-\epsilon} + \int_{\epsilon}^R \right) \frac{1}{x} \frac{1}{2i} \left( e^{ix + ip x} - e^{-ix + ipx} \right) \, dx$$
Note that if we keep ##\sin(x)## intact, we can integrate through ##0## because ##\sin(x)/x## has no singularity at ##x \to 0##: this function can be defined as ##1## when ##x = 0.## However, when we split up ##\sin(x)## into exponentials, we need to keep away from ##x=0##, hence the ##(-\epsilon,+\epsilon)## exclusion interval.Let ## I_1 ## be the integral containing ##e^{ix + ip x}## and ##I_2## be the other one, containing ##e^{-ix + ipx}##.

(1) Evaluation of ##I_1.## Since we are assuming ##p \geq 0##, we can complete ##I_1## in the upper ##z-##plane (that is, take ##z = x + iy, \:y \geq 0##). Define the points ##A = \epsilon +i0##, ##B = R + i0##, ##C = R + iR##, ##D = -R + iR##, ##E = -R + i0## and ##F = -\epsilon + i0##. Take the closed contour ##\gamma_1 ## equal to ##A \to B \to C \to D \to E \to F \to A##, where the last part is clockwise around the semicircle of radius ##\epsilon## in the upper half-plane. Of course, for ##f_1 = \frac{1}{2i z} e^{i(1+p)z}## we have ##\int_{\gamma_1} f_1 \, dz = 0,## as you note. Now
$$\int_B^C f_1 = \int_0^R \frac{1}{2 i (R+iy)} e^{i(1+p) (R + iy)} i \, dy,$$ so
$$\left|\int_B^C f_1 \right| \leq \int_0^R \frac{1}{2R} e^{-(1+p) y} dy \to 0 \; \text{as} \; R \to \infty.$$
Similarly, ##\int_C^D f_1 \to 0## and ##\int_D^E f_1 \to 0## as ##R \to \infty.## So, in the ##R \to \infty## limit we have ##\int_{-\infty \to F} f_1 + \int_{F \to A} f_1 + \int_{A \to \infty} f_1 = 0.## That is, ## ( \int_{-\infty \to F} + \int_{A \to \infty} ) f_1 = - \int_{F \to A} f_1,## where the last integral is clockwise around the small semi-circle. It is easy to compute in the limit ##\epsilon \to 0+.##

(2) Evaluation of ##I_2.## For the case of ##0 < p < 1## we can complete ##f_2 = \frac{1}{2iz} e^{-ix + ipx}## in the lower half-plane, much as was done above, and then compute the limiting integral. However, if ##p > 1## we must complete ##f_2## in the upper half-plane, and compute the limiting integral.

For the case of ##p = 1## we can completer ##f_1## in the upper half-plane and ##f_2## in the lower half-plane.
To be honest, it is a really nice answer, but it does not follow up my original questions, plus I am having a hard time trying to picture the contour you are describing, at least the part from C and D. Still thanks.

Santilopez10 said:
To be honest, it is a really nice answer, but it does not follow up my original questions, plus I am having a hard time trying to picture the contour you are describing, at least the part from C and D. Still thanks.

The points ##A## to ##F## were described in detail, so drawing a sketch would be the easiest way to see what is happening. The segment ##B \to C## is vertically up, ##C \to D## is horizontal from right to left, etc. You are going around four sides of a ##2R \times R## rectangle with a small gap in the middle of the bottom side.

As for your original question: my integral of ##f_1## over ##B \to C \to D \to E## is the same as the integral of ##f_1## over the semicircle ##B \to E##, but I find that getting bounds on vertical and horizontal segments is much, much easier than trying to get a bound on the semicircle; that is why I avoided the semicircle. Furthermore, I really do believe you must split up the ##\sin x## into its exponential parts and then (at least for ##0 \leq p < 1##) complete one part in the upper half-plane and the other part in the lower half-plane. The fact that when ##p > 1## you need to complete both parts in the upper half-plane is the reason you get ##0## in this case.

Anyway: how can you tell which method to use? Sometimes semicircles work like a snap and rectangles are hard, but other time semicircles are hard and rectangles are easy. Basically, the way I used to do these things back in the stone age when I was a grad student was to try one thing, and if it did not seem to be working easily, then stop wasting time and go on to try something else. After doing many such problems one begins to develop a "feeling" for the most likely prospect.

Finally: my remarks about the case ##p=1## were not useful. I think it is better to just write ##e^{ix} = \cos x + i \sin x##, and look at the functions ##\sin x \cos x / x## and ##\sin^2 x /x ## separately, noting their "parity". That leads to a lot of simplification.

I think this can be dealt without considering at all contour integrals over the complex plane, just considering integrals over the real line, replacing ##e^{ipx}=\cos{px}+i\sin{px}## and calculating the real and imaginary part of the integral. It is easy to see that the imaginary part is zero because the function is odd.

Delta2 said:
I think this can be dealt without considering at all contour integrals over the complex plane, just considering integrals over the real line, replacing ##e^{ipx}=\cos{px}+i\sin{px}## and calculating the real and imaginary part of the integral. It is easy to see that the imaginary part is zero because the function is odd.
but it is a complex analysis exercise...

If someone can provide an answer which follows the semicircle contour, I would be so glad.

Santilopez10 said:
If someone can provide an answer which follows the semicircle contour, I would be so glad.

I looked more carefully at your semicircle contour analysis, and was impressed that you obtained a bound by using concavity of the function ##\sin w## on ##w \in [0,\pi/2].## Well done, very clever. You get a useful bound: for any ##p > -1## your bounding function
$$\frac{1-e^{-(p+1)R}}{(p+1) R}$$ goes to ##0## as ##R \to \infty##, so the semicircular part goes away in the limit (at least if ##p > -1##).

What it all amounts to is that you have essentially completed the calculation for the ##e^{+ix}## part of ##\sin x.## Now you need to do it again, for the ##e^{-ix}## part. For some reason you seem reluctant to acknowledge this. I suspect that you will never be successful as long as you resist doing this. You are not getting an answer of ##0## when ##p > 1##, and you are not getting an answer of ##\pi## when ##0 \leq p < 1.## In fact, you are getting a purely imaginary answer rather than the true, real one.

In other words, for ##p > -1## you are getting the Fourier transform ##F(p)## of ##e^{ix}/x,## not of ##\sin x/x .##

Last edited:
Ray Vickson said:
I looked more carefully at your semicircle contour analysis, and was impressed that you obtained a bound by using concavity of the function ##\sin w## on ##w \in [0,\pi/2].## Well done, very clever. You get a useful bound: for any ##p > -1## your bounding function
$$\frac{1-e^{-(p+1)R}}{(p+1) R}$$ goes to ##0## as ##R \to \infty##, so the semicircular part goes away in the limit (at least if ##p > -1##).

What it all amounts to is that you have essentially completed the calculation for the ##e^{+ix}## part of ##\sin x.## Now you need to do it again, for the ##e^{-ix}## part. For some reason you seem reluctant to acknowledge this. I suspect that you will never be successful as long as you resist doing this. You are not getting an answer of ##0## when ##p > 1##, and you are not getting an answer of ##\pi## when ##0 \leq p < 1.## In fact, you are getting a purely imaginary answer rather than the true, real one.

In other words, for ##p > -1## you are getting the Fourier transform ##F(p)## of ##e^{ix}/x,## not of ##\sin x/x .##
okey I am starting to get what you mean, so for different values of P I have to close the contour in different parts so that the extra stuff goes to 0 as R approaches infinity am I right? I find it strange that I can not use my ##e^{iz}## substitution instead of writing sine as exponentials.

Santilopez10 said:
okey I am starting to get what you mean, so for different values of P I have to close the contour in different parts so that the extra stuff goes to 0 as R approaches infinity am I right? I find it strange that I can not use my ##e^{iz}## substitution instead of writing sine as exponentials.

Well, you want to extend ##\sin x## into the complex plane. The definition of ##\sin z## is
$$\sin z = \frac{1}{2i} (e^{iz} - e^{-iz} ) \hspace{4ex}(1)$$ so that is how it must be extended.

Actually, another definition of ##\sin z## is
$$\sin z = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots \hspace{4ex}(2)$$
but one can show that (1) and (2) are equivalent.

Whether or not you need to use different contours for the two parts depends on the magnitude of ##p##. If ##0 < p < 1## you need two different contours, but if ##p > 1## you need only one contour. Try it and see!

1. What is a complex integral problem?

A complex integral problem is a mathematical problem that involves finding the value of a complex integral, which is essentially a type of line integral in the complex plane. It involves integrating a complex-valued function along a curve in the complex plane.

2. How is a complex integral problem different from a regular integral problem?

A complex integral problem differs from a regular integral problem in that it involves integrating a complex-valued function along a curve in the complex plane, rather than a real-valued function along a curve in the real plane. This requires the use of complex analysis techniques and the Cauchy-Riemann equations.

3. What are some common techniques used to solve complex integral problems?

Some common techniques used to solve complex integral problems include the Cauchy integral theorem, the Cauchy integral formula, and the residue theorem. These techniques involve using contour integration and the properties of analytic functions in the complex plane.

4. What are some real-world applications of complex integral problems?

Complex integral problems have many real-world applications, particularly in physics and engineering. They are used to solve problems related to electric fields, fluid flow, heat transfer, and many other physical phenomena. They are also used in signal processing and image analysis.

5. What are some tips for solving complex integral problems?

Some tips for solving complex integral problems include choosing an appropriate contour for integration, using the Cauchy-Riemann equations to simplify the integrand, and being familiar with the properties of analytic functions. It is also important to carefully evaluate any singularities or poles in the complex plane.

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