Solving Complex Integrals: Eliminating isin2bx with Analytic Functions

[Daniiel]

Hey,

Got stuck studying complex stuff again, I am trying to find out how i can get rid of the isin2bx in my result, here is the question

[PLAIN]http://img832.imageshack.us/img832/3199/unledkcv.jpg

The integral of e^(-z^2) = 0 as C is a closed curve and e^(-z^2) is analytic

So first parametrizing the rectagle, letting the right side by C1, top C2, then C3, C4, where the paths C2 and C4 are longer.

So for C1
p(t) = R+it, for 0<=t<=b, then d(p(t)/dt =i
so its integral betweel b and 0 e^-(R+it)^2 idt = blah (expand the power) which tends to 0 as R tends to infinity.

and so on for each path

Adding each path together i get

sqrt(Pi) - 2e^(-b^2) int(between zero and R) e^(-x^2) (cos(2bx)-isin(2bx)) = 0

so (sqrt(pi)e^(b^2))/2 = int(between zero and R) e^(-x^2) (cos(2bx)-isin(2bx)), let R tend to infinity

Im trying to work out how can i get rid of the isin(2bx)?

I'm also not sure what it means by "Why is the answer sensible" what do you guys think it means?

Thanks in advanced



[itex] [/itex]

 

[susskind_leon]

My guess would be to write
$$\cos(2bx)=\frac{e^{i2bx}+e^{-i2bx}}{2}$$
So since you already got the expression for $\int_0^\infty e^{-x^2}e^{-i2bx}dx$, you just replace b by -b to get the expression for $\int_0^\infty e^{-x^2}e^{i2bx} dx$, then add that sh*t together and divide by 2. As for "why is your answer sensible", wel... it depends on what they would like to hear. My guessing is that a) for b=0, you get the known value for the Gaussian integral b) the larger b, the more your function oscillates in the "important" region around x=0, so the integral will be smaller for larger b, yet it will always be positive.

 

[Daniiel]

I'm not really sure what you mean,

If you write cos and sin in there exponential forms you'll get the first integral you wrote, but how does that remove the sin to provide the e(-x^2)cos(2bx) required?

 

[susskind_leon]

Remember Euler's formula
$$e^{ix} = \cos(x) + i \sin(x)$$
and for the negative argument, you get
$$e^{-ix} = \cos(x) - i \sin(x)$$
You have calculated
$$e^{-x^2} (\cos(2bx)-i\sin(2bx) )=e^{-x^2} e^{-i2bx}$$

 

[Daniiel]

Yea I understand that, but I need to turn that into just e^(-x^2)cos(2bx) so I can answer the question.

It asks for the integral of e^(-x^2)cos(2bx), and i have
the integral of f e^(-x^2)[cos(2bx)-isin(2bx)] or e^(-x^2)e^(-i2bx) how ever you want to write it

oooo

Can i just say that there is no imaginary component, on the left side (sqrt(pi)e^(-b^2)) so the imaginary part on the right must be zero.

But then that infers b = npi for any integer n

 

[susskind_leon]

You got an expression for
(sqrt(pi)e^(b^2))/2 = int(between zero and R) e^(-x^2) (cos(2bx)-isin(2bx))

Now just b into -b to get an expression for
(sqrt(pi)e^(b^2))/2 = int(between zero and R) e^(-x^2) (cos(2bx)+isin(2bx))

(remember the evenness of cosine and oddness of sine. Now add that together and divide by 2.

 

[Daniiel]

I know the answer is this

integral between 0 and infinity e^(-x^2)cos(2bx) = e^(b^2)sqrt(pi)/2

and in other words, trying to get rid of the sin so I can represent the answer correctly

I really don't see how what your doing gets that

 

[susskind_leon]

Seriously, I don't understand your problem... You know
$$\int_0^\infty e^{-x^2} (\cos(2bx) - i \sin(2bx)) = \frac{\sqrt{\pi}e^{-b^2}}{2}$$
Now you can follow that
$$\int_0^\infty e^{-x^2}(\cos(2bx) + i \sin(2bx)) = \frac{\sqrt{\pi}e^{-b^2}}{2}$$
Adding those two equations gives you
$$\int_0^\infty e^{-x^2}(2\cos(2bx)) = 2 \frac{\sqrt{\pi}e^{-b^2}}{2}$$
which ultimately lease to
$$\int_0^\infty e^{-x^2}\cos(2bx) = \frac{\sqrt{\pi}e^{-b^2}}{2}$$
what's unclear?

 

[Daniiel]

ahahah, sorry

the whole time i thought you were saying add cos(2bx) to -sin(2bx)

thanks haha, sorry again

 

[susskind_leon]

no worries mate ;)
glad you finally figured it out :)