Solve Difficult Integral: Analytic Approximations Welcome

  • Thread starter Rednas
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In summary: Let's hope someone will take the time to work this out, because it really looks nasty!Probably, unfamiliarity with the advanced topic is creating the problem. But, as I know, this particular integral is of Fourier Integral Theorem, and it solved from the left to the right i.e. formal tactics of a regular double integration not necessarily followed. This means, this integral does not generates a surface of the solid, but rather simplifies the wave equation from amplitude to frequency domain. Or the same analogy holds also for any sinusoidal signal wave. So, this integral does generate a surface of the solid, but it's a tedious job to get there.
  • #1
Rednas
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I fear that this one is really hard, if not impossible, but an analytic answer would be way more usefull than a numerical one. Who can help me in the right direction?




[itex]\int_0^{arccos(a)} d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2})(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]

with 0<a<1 and the phi integral only over positive values of the squareroot

Approximations for y=0 and a small are also welcome.

This integral comes from the double integral [itex]\int_0^{\infty} dk\int_0^{2\pi}d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2}} e^{i k(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]
 
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  • #2
i don't understan anything of what you wrote.
could you post somthing more readible...

use the buttons if you don't understand tex code.

ciao
marco
 
  • #3
Marco_84 said:
i don't understan anything of what you wrote.
could you post somthing more readible...

use the buttons if you don't understand tex code.

ciao
marco

Yeah i agree with u! i am staring at it for a few minutes but it is quite hard to dechiper, quite ambiguous!
 
  • #4
sutupidmath said:
Yeah i agree with u! i am staring at it for a few minutes but it is quite hard to dechiper, quite ambiguous!
Ambiguous? I don't think that word means what you think it means. :-p The expression is written in the syntax of Mathematica. (and I'm pretty sure it's syntactically correct)
 
  • #5
Hurkyl said:
Ambiguous? I don't think that word means what you think it means. :-p The expression is written in the syntax of Mathematica. (and I'm pretty sure it's syntactically correct)
Yeah i guess! I just wanted to say that it is hard to read what the OP posted, but i guess i said the wrong way!
 
  • #6
Posted something more readable. Wasn't aware of the Tex possibilities at first, so I used the Mathematica syntax.
 
  • #7
Rednas said:
I fear that this one is really hard, if not impossible, but an analytic answer would be way more usefull than a numerical one. Who can help me in the right direction?




[itex]\int_0^{arccos(a)} d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2})(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]

with 0<a<1 and the phi integral only over positive values of the squareroot

Approximations for y=0 and a small are also welcome.

This integral comes from the double integral [itex]\int_0^{\infty} dk\int_0^{2\pi}d\phi \frac{cos(\phi)}{(cos(\phi)+\sqrt{cos^2(\phi) - a^2}} e^{i k(x \sqrt{cos^2(\phi)-a^2}+y sin(\phi) + z + x_0 cos(\phi) )}[/itex]

Let's hope someone will take the time to work this out, because it really looks nasty!
 
  • #8
Probably, unfamiliarity with the advanced topic is creating the problem. But, as I know, this particular integral is of Fourier Integral Theorem, and it solved from the left to the right i.e. formal tactics of a regular double integration not necessarily followed. This means, this integral does not generates a surface of the solid, but rather simplifies the wave equation from amplitude to frequency domain. Or the same analogy holds also for any sinusoidal signal wave. The original question was not that complex, when all variables get converted into spherical or cylindrical parametric form, it looks quite messy, but in reality, it is only a tedious job, not genuine.
 

Related to Solve Difficult Integral: Analytic Approximations Welcome

1. What is an analytic approximation?

An analytic approximation is a method used to estimate the value of a complicated integral by breaking it down into simpler, more manageable parts. It involves using mathematical techniques and formulas to find a close, but not exact, solution to the integral.

2. Why is it important to solve difficult integrals?

Difficult integrals often arise in many scientific and engineering applications, and being able to solve them accurately is crucial for obtaining precise results. Analytic approximations provide a way to handle these integrals without having to resort to numerical methods, which can be time-consuming and less accurate.

3. What are some common techniques used in analytic approximations?

Some common techniques used in analytic approximations include integration by parts, change of variables, and series expansions. Each of these techniques can be applied to different types of integrals, depending on their complexity.

4. How do I know if my analytic approximation is accurate?

The accuracy of an analytic approximation can be evaluated by comparing it to the exact solution, if it is known. Otherwise, it can be compared to numerical approximations or validated by testing it with different inputs and checking if the results are within acceptable error margins.

5. Are there any limitations to using analytic approximations?

Yes, there are some limitations to using analytic approximations. They may not work for all types of integrals, especially those with highly oscillatory or rapidly changing functions. Additionally, the accuracy of the approximation may decrease as the complexity of the integral increases.

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