Difficult cosh integral using Leibniz rule?

In summary, the conversation is about a student seeking help with a specific definite integral problem in quantum mechanics using the WKB approximation. The integral includes a positive constant, and the student has tried different approaches without success. They have asked their professor for guidance, who suggested the Leibnitz rule, but the student is unsure how it would apply in this case. They clarify that the integral is definite with specific limits, and the student has attempted to solve it but has only been successful for a=1. The student is unsure if the integral is solvable and mentions not having knowledge about the gamma function or airy functions.
  • #1
Kavorka
95
0
I was wondering if I could get some pointers on how to at least start on this. In quantum mechanics we are using the WKB approximation, and we end up with a definite integral that looks like this:

∫(1 - a(cosh(x))-2)1/2 dx = ∫(1/cosh(x)) (1 - a(cosh(x))2)1/2 dx

where a is a positive constant. I've tried everything I can think of to no avail, the answer on wolfram isn't pretty but it seems like if I can figure out what process to use I could reach it eventually. I asked the professor and he suggested Leibnitz rule, but not sure how differentiation under the integral sign would help here.
 
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  • #2
Kavorka said:
definite integral

But it does not have limits on it. Do you mean indefinite ? I am asking because some indefinite integrals are easy to evalute with limits.

Are you just interested in final result ?
 
  • #3
Buffu said:
But it does not have limits on it. Do you mean indefinite ? I am asking because some indefinite integrals are easy to evalute with limits.

Are you just interested in final result ?

From how my professor was describing it it seemed like the limits wouldn't be too helpful, but yes it is a definite integral from -cosh-1(a0.5 to cosh-1(a0.5 (inverse hyperbolic cosh, not cosh^-1 )
 
  • #4
Kavorka said:
From how my professor was describing it it seemed like the limits wouldn't be too helpful, but yes it is a definite integral from -cosh-1(a0.5 to cosh-1(a0.5 (inverse hyperbolic cosh, not cosh^-1 )

I tried to integrate it and I was unable to do so except for ##a = 1## (which you can do easily). I think it is not integratable but I am no expert, I don't have any clue about gamma function, airy functions ...
 

1. How do you use the Leibniz rule to solve difficult cosh integrals?

The Leibniz rule, also known as the product rule, is used to differentiate a product of two functions. In the case of difficult cosh integrals, the Leibniz rule can be used to simplify the integral into a more manageable form. This is achieved by splitting the integral into two separate integrals using the Leibniz rule, and then solving each integral individually.

2. What is the general form of the Leibniz rule?

The Leibniz rule states that the derivative of a product of two functions is equal to the first function times the derivative of the second function, plus the second function times the derivative of the first function. In mathematical notation, this can be expressed as (f * g)' = f' * g + f * g'.

3. Are there any special cases where the Leibniz rule cannot be applied?

Yes, the Leibniz rule cannot be applied if the functions involved are not differentiable, or if the product of the functions is not integrable. In these cases, other methods may need to be used to solve the integral.

4. How can the Leibniz rule be used to simplify cosh integrals?

The Leibniz rule can be used to simplify cosh integrals by breaking the integral into two separate integrals, and then using the Leibniz rule to find the derivatives of the functions involved. This can often result in a simpler form of the integral that is easier to solve.

5. Can the Leibniz rule be used to solve other types of integrals?

Yes, the Leibniz rule can be used to solve other types of integrals, such as integrals involving trigonometric functions or exponential functions. It can also be extended to higher dimensions for multivariate functions. However, it is important to note that the Leibniz rule may not always be the most efficient method for solving integrals, and other techniques may be more suitable depending on the specific integral at hand.

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