Evaluate integral by using residues: complex variables

[sara_87]

Homework Statement

evaluate:

\int_{-\infty}^{\infty}e^{-ax^2}cos(bx)dx

Homework Equations

The Attempt at a Solution

since \cos(bx) is the real part of: e^(b*x*i),
we can rewrite the integral as:

Real[\int_{-\infty}^{\infty}e^{bxi-ax^2}dx]

but now I am stuck because I don't know which contour to choose and i don't know where the poles(singularities) are.

thank you.

 

[Dick]

Don't do it that way. Write cos(bx) as (exp(ibx)+exp(-ibx))/2.

 

[sara_87]

ok, this will give as the integrand:

\frac{1}{2}(e^{ibx-ax^2}+e^{-ibx-ax^2})

I should have mentioned that this question requires us to use contour integration using residues.

 

[Count Iblis]

doing what Dick is suggesting is in fact equivalent as using the contour integration method. After completing the square and changing variables, the integral is no longer along the real axis. If you consider the same integral as you get here but now along the real axis, then that integral has the same value by Cauchy's theorem. This last step is often omitted, people often don't even notice that after completing the square the integral is no longer along the real axis.

 

[sara_87]

after completing the square the integral is no longer along the real axis.

I don't understand. How do complete the square here? and what do you mean: 'the integral is no longer along the real axis' ?

 

[Count Iblis]

I don't understand. How do complete the square here? and what do you mean: 'the integral is no longer along the real axis' ?

Then, let's forget this for the moment (you'll only understand it if you already know how to do the problem) and simply consider the following contour integral:

I\left(R,p\left)=\oint_{C\left(R,p\right)}\exp\left(-a z^{2}\right)dz

where C\left(R,p\right) is the contour from -R to R and then from there to R + i p, from there to -R + i p and then back to the starting point -R.

Questions for you:

a) What do the theorems about contour integration tell you about the value of I\left(R,p\right)?

b) In the limit R\to\infty what happens to the parts of the contour integral from R to R + i p and from -R + i p to -R?

c) From a) and b) what can you conclude about the integral:

\int_{-\infty}^{\infty}\exp\left[-a\left(x+ip\right)^2\right]dx

d) Can you choose p in the above integral so that you get your desired integral up to some factor?

 

[sara_87]

What do you mean when you say from R to R+ip? is this a straight line going up or is this a semi-circle?
a) which theorem? you mean Cauchy's theorem? the one that says that I=0 provided the function is analytic

 

[Count Iblis]

What do you mean when you say from R to R+ip? is this a straight line going up or is this a semi-circle?
a) which theorem? you mean Cauchy's theorem? the one that says that I=0 provided the function is analytic

Yes, a straight line. You can imagine p to be some fixed number while R is supposed to go to infinity in the end.

And yes, Cauchy's theorem implies that I(R,p)=0 because exp(-z^2) is analytic.

 

[sara_87]

for part b)
as R tends to infinty, R+ip will still stay the same becasue it's just a verical line. I have a feeling that R+ip should tend to 0 but i don't know why.

 

[Count Iblis]

for part b)
as R tends to infinty, R+ip will still stay the same becasue it's just a verical line. I have a feeling that R+ip should tend to 0 but i don't know why.

Your feeling is correct, so you need to investigate this in more detail. The integral from R to R + i p can be written as:

\int_{0}^{p}\exp\left[-a\left(R + i y\right)^2\right]i dy

How does this integral behave in the limit of R to infinity?

 

[sara_87]

oh yeah, of course. when R tends to infinity, the integrand becomes very small (e^(-infinity)) and therefore tends to 0.

for c) does this mean that:
from -R to R: integral=0 because of Cauchy's theorem. ?
from R to R+ip = from -R+ip to -R= 0 because as R tends to infinty, the intergand tends to zero.
but what about from R+ip to -R+ip ?

 

[Count Iblis]

oh yeah, of course. when R tends to infinity, the integrand becomes very small (e^(-infinity)) and therefore tends to 0.

for c) does this mean that:
from -R to R: integral=0 because of Cauchy's theorem. ?
from R to R+ip = from -R+ip to -R= 0 because as R tends to infinty, the intergand tends to zero.
but what about from R+ip to -R+ip ?

Yes, the integral from R to R + i p tends to zero and so does the integral from -R + i p to -R, but you have to be able to prove that more formally, let's skip that step for the moment.

Now Cauchy's theorem makes a statement about the whole contour integral, not just a few parts of it. The entire contour integral is zero, while two parts of it tend to zero. So, what can you conclude about the sum of the two other parts?

 

[sara_87]

for the two other parts:
they are also analytic? is it 0? but i don't know why.

 

[Count Iblis]

What does Cauchy's theorem say about the integral along a closed path of an analytic function such as exp[-az^2]?

In this case, the closed path consists of the four line segments.

 

[sara_87]

It says that if f is analytic an in a closed contour then the integral=0

you mean even \frac{1}{2}(e^{ibx-ax^2}+e^{-ibx-ax^2})
can be applied to cauchys theorem?

 

[Count Iblis]

It says that if f is analytic an in a closed contour then the integral=0

you mean even \frac{1}{2}(e^{ibx-ax^2}+e^{-ibx-ax^2})
can be applied to cauchys theorem?

Yes, provided you integrate it along a closed contour.

So, do you see now that the sum of the four integrals along the line segments is always zero, while the integral along two of the line segments from R to R + i p and from -R + i p to -R tend to zero, implying that the integral from -R to R plus the integral from
R + i p to -R + i p tends to zero in the limit R to infinity?

 

[sara_87]

I see. just one last question. I understand that the vertical lines have integral 0 as R tends to infinity, but why does this 'imply' that the integral of the horizontal lines = 0?

 

[Count Iblis]

I see. just one last question. I understand that the vertical lines have integral 0 as R tends to infinity, but why does this 'imply' that the integral of the horizontal lines = 0?

The sum of the two integrals over the horizontal lines tends to zero because, by Cauchy's theorem, the sum of all four integrals is zero (the sum of all four is the contour integral).

 

[sara_87]

oh right i see.
Im sorry, in the question i posted i said the integral goes from -\infty to +\infty but it should be from 0 to \infty. In this case, do we take the same contour as we did before?

 

[Count Iblis]

oh right i see.
Im sorry, in the question i posted i said the integral goes from -\infty to +\infty but it should be from 0 to \infty. In this case, do we take the same contour as we did before?

Because the integrand is an even function, the integral from zero to infinity is half the integral from minus to plus infinity. So, you can take the same contour and compute the integral from minus to plus infinity and then divide that by 2.

 

[sara_87]

''d) Can you choose p in the above integral so that you get your desired integral up to some factor? ''
so does this mean i have to equate and find an expression for p? I mean do i have to find an expression for p from the following:

\exp\left[-a\left(x + i y\right)^2\right]i dy=\frac{1}{2}(e^{ibx-ax^2}+e^{-ibx-ax^2})


?

 

[Count Iblis]

If you've done c) correctly, you should have found that Cauchy's theorem and the fact that the two integrals from R to R + i p and the one from -R +i p to -R tend to zero, imply that:

\int_{-\infty}^{\infty}\exp\left[-a\left(x+i p\right)^2\right]dx=\int_{-\infty}^{\infty}\exp\left(-a x^2\right)dx

The integral on the right hand side is

\int_{-\infty}^{\infty}\exp\left(-a x^2\right)dx=\sqrt{\frac{\pi}{a}}

All you have to do now is expand the square in the exponential on the left hand side:

(x+ip)^2=x^2 + 2 i p x -p^2

So

\exp\left[-a\left(x+ip\right)^2\right]=\exp\left(-ax^2 -2iapx\right)\exp\left(ap^2\right)

You know that the integral of exp(-ax^2)cos(bx) is the real part of the integral of
exp(-ax^2 + i b x). So, by choosing p appropriately in the above expression, you can make sure that the term linear in x is i b.

 

[sara_87]

ok, so how about: p= -b/(2a)
?

 

[Count Iblis]

ok, so how about: p= -b/(2a)
?

Looks good!

What do you then obtain for the integral?

 

[sara_87]

We want the real part of:

\int_0^{\infty} \exp(-ax^2+ibx)

so let A= \int_0^{\infty} \exp(-a(x-\frac{bi}{2a})^2)

we need:

Ae^(b^2/4a)

so we need:

\exp(b^2/4a)\int_0^{\infty} \exp(-ax^2+ibx)dx

right?

 

[Count Iblis]

We want the real part of:

\int_0^{\infty} \exp(-ax^2+ibx)

so let A= \int_0^{\infty} \exp(-a(x-\frac{bi}{2a})^2)

we need:

Ae^(b^2/4a)

so we need:

\exp(b^2/4a)\int_0^{\infty} \exp(-ax^2+ibx)dx

right?

Yes, and you know all along (using Cauchy's theorem) that the real part of the last integral (including the factor exp(b^2/4a)) is equal to the known integral of exp(-ax^2). So, you've found the answer.

 

[sara_87]

ah right...i see :)

so the answer is

\sqrt{\pi/a}\exp(-b^2/(4a))

but to get the integral from 0 to infinity, we divide this by 2.

right?

 

[Count Iblis]

ah right...i see :)

so the answer is

\sqrt{\pi/a}\exp(-b^2/(4a))

but to get the integral from 0 to infinity, we divide this by 2.

right?

Correct!

I'll explain tomorrow how you can compute the same integral using a slightly different contour.

 

[sara_87]

Cool!

What the other method?

:)

 

[Count Iblis]

Cool!

What the other method?

:)

Consider the following contour:

We start at z = 0 and move along the real axis to x = R, from there in the imaginary dierection to z = R + i p, and then parallel to the real axis to z = i p oin the imaginary axis and then back to z = 0.

Cauchy's theorem says that
\oint_{C}\exp\left(-az^{2}\right)dz=0
where C is the above defined contour. If you then write out the contour integral in terms of the contributions of the 4 integrals along the four lines and take the limit R\to\infty, you get:

\int_{0}^{\infty}\exp\left(-ax^{2}\right)dx -\int_{0}^{\infty}\exp\left[-a\left(x+i p\right)^{2}\right]dx-i \int_{0}^{p}\exp\left(ay^{2}\right)dy=0

You then get the same result you obtained earlier by taking the real part of this equation. if you take the imaginary part you get an identity for the integral with cos replaced by sin.