Finding the Correct Solution to an Integral in Quantum Mechanics

[jbowers9]

Homework Statement

I recently tried to do the following integral:
a_n = ∫√(2/a) sin(n∏x/a) cosh(x) dx
x=0 to x=a

Homework Equations

a_n = ∫√(2/a) sin(βx) cosh(x) dx
β = n∏/a
sin(βx) = ½i(e^iβx – e^-iβx)
cosh(x) = ½(e^x + e^-x)

The Attempt at a Solution

a_n = ¼ i √(2/a)∫ (e^iβx – e^-iβx) (e^x + e^-x)

after all is said and done, I get;

a_n = √(2/a)[(a²sin(n∏)sinh(a) – acos(n∏)cosh(a) + n∏a)/(n²∏² + a²)]


The text, “Quantum Mechanics Demystified”, however, gets;

a_n = √(2/a)[a(n∏cos(n∏)cosh(a) + sin(n∏)sinh(a))/( n²∏² + a²)]

Which is correct? And why?

 

[Mark44]

Homework Statement

I recently tried to do the following integral:
a_n = ∫√(2/a) sin(n∏x/a) cosh(x) dx
x=0 to x=a

Homework Equations

a_n = ∫√(2/a) sin(βx) cosh(x) dx
β = n∏/a
sin(βx) = ½i(e^iβx – e^-iβx)
cosh(x) = ½(e^x – e^-x)

The Attempt at a Solution

a_n = ¼ i √(2/a)∫ (e^iβx – e^-iβx) (e^x – e^-x)

after all is said and done, I get;

a_n = √(2/a)[(a²sin(n∏)sinh(a) – acos(n∏)cosh(a) + n∏a)/(n²∏² + a²)]


The text, “Quantum Mechanics Demystified”, however, gets;

a_n = √(2/a)[a(n∏cos(n∏)cosh(a) + sin(n∏)sinh(a))/( n²∏² + a²)]

Which is correct? And why?

Your approach is the one I would take, so here is what I would do:

Check your work to see if you can find any errors.
Take the derivative of your result. Do you get the integrand?
Take the derivative of the book's result. Do you get the integrand?
If the answers to both questions are yes, the two antiderivatives are equal or differ by a constant.
If one answer is yes and the other is no, the result from the "yes" answer is almost surely correct and the other is incorrect. It's even possible that the answer in the book is wrong.

 

[gabbagabbahey]

Homework Statement

I recently tried to do the following integral:
a_n = ∫√(2/a) sin(n∏x/a) cosh(x) dx
x=0 to x=a

The Attempt at a Solution

I get;

a_n = √(2/a)[(a²sin(n∏)sinh(a) – acos(n∏)cosh(a) + n∏a)/(n²∏² + a²)]

If you've written the Integral correctly, then your solution is closer than the one from the text; it should have an aSin(...) term instead of an a²Sin(...)...Of course, if n is an integer then the sin (n*pi) term is zero and cos(n*pi)=(-1)ⁿ.

So, are you sure you are evaluating the correct integral?

 

[gabbagabbahey]

Your approach is the one I would take, so here is what I would do:

Check your work to see if you can find any errors.
Take the derivative of your result. Do you get the integrand?
Take the derivative of the book's result. Do you get the integrand?
If the answers to both questions are yes, the two antiderivatives are equal or differ by a constant.
If one answer is yes and the other is no, the result from the "yes" answer is almost surely correct and the other is incorrect. It's even possible that the answer in the book is wrong.

Usually these are good strategies for checking a solution, but in this case the integral is a definite integral, so differentiating the textbook's solution will naturally give zero.

 

[jbowers9]

I made an error transcribing the above and corrected the cosh(x) term. I've redone it 3 times and still get my results.