How to find the expectation value of cos x

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1. Dec 7, 2018

DEEPTHIgv

1. The problem statement, all variables and given/known data
If x is a continuous variable which is uniformly distributed over the real line from x=0 to x -> infinity according to the distribution f (x) =exp(-4x) then the expectation value of cos 4x is?

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2. Relevant equations
the expectation value of any function g is, <g>=∫Ψ*gΨdx

3. The attempt at a solution

Here,
Ψ=f(x)
since f is independent of time
∴Ψ*=f(x)
now, for <cos 4x>=∫exp(-4x)cos(4x)exp(-4x)dx
let <cos4x>=I=∫cos(4x)exp(-8x)dx
rest of the solution if in the picture attached
my main doubt is how to put the limit infinity in sin and cos

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Last edited by a moderator: Dec 7, 2018
2. Dec 7, 2018

Delta2

The limits $\lim_{x\rightarrow \infty}\cos{4x}$, $\lim_{x\rightarrow \infty}\sin{4x}$ do not exist, however here you have the limits

$\lim_{x\rightarrow \infty}e^{-8x}\cos{4x}$ , $\lim_{x\rightarrow \infty}e^{-8x}\sin{4x}$ which are zero, because $lim_{x\rightarrow \infty}e^{-8x}=0$ and the functions $\sin{4x}, \cos{4x}$ are bounded. (It is a well known lemma that the limit of the product of a bounded function with another function that has limit zero, is also zero).

3. Dec 7, 2018

ehild

The probability distribution function is not the same as the wavefunction in QM.

4. Dec 8, 2018

haruspex

This does not read like a quantum theory question to me, just straight probability. But the given f(x) is not a valid pdf; its integral over the full range is not 1. Seems like it should be 4e-4x.
Anyway, to answer your question, all you care about at infinity is that the trig functions are bounded. Look at the other factor.

Beaten by Δ2 and ehild because I spent time figuring out the answer, and I do not believe it is 1/2. Also, there is still the problem that the given pdf is not valid.

Edit: the reason I got a different answer is that I used cos(x), as stated in the title, instead of cos(4x), as stated in post #1.

Last edited: Dec 8, 2018
5. Dec 8, 2018

Delta2

That was my concern also, but my subconscious thought was that he meant wave function, though the problem statement clearly states probability density function.

So the initial integral to start with would be $I=\int_0^{\infty}4e^{-4x}\cos{4x}dx$.

6. Dec 8, 2018

haruspex

Except, as I posted, it fails the totality of 1 criterion.

7. Dec 8, 2018

Delta2

Yes it should have been with a factor 4 in front (as you already said), probably a typo by the OP.

8. Dec 8, 2018

haruspex

You may find it a bit simpler to spot that a likely integral of $e^{-\lambda x}\cos(\mu x)$ is the form $e^{-\lambda x}(A\cos(\mu x)+B\sin(\mu x))$ then differentiate that to find A and B.

9. Dec 8, 2018

DEEPTHIgv

Could you please explain or suggest a source from which I can understand the difference.

10. Dec 8, 2018

DEEPTHIgv

It is possible.
The above question is from a previous year entrance test.

11. Dec 8, 2018

DEEPTHIgv

Isn't the expectation value supposed to be ∫f* cos4x f dx

12. Dec 8, 2018

DEEPTHIgv

I have posted the question just as it was given.

13. Dec 8, 2018

Delta2

That is correct if $f$ is given as a wave function. However the problem states that $f$ is a probability density function hence the expectation value of the variable $g(x)$ is simply $\int g(x)f(x)dx$.

To unify the two cases, when $f$ is the wave function then the probability density function is $|f|^2=f^{*}f$

Last edited: Dec 8, 2018
14. Dec 8, 2018