Could someone help me to evaluate the following integral? I'm having trouble getting a result using a corollary of Cauchy's Integral Formula. When I break the trig functions down to their exponential forms, it gives me two integrals that are ok to solve using that corollary but also another two integrals which seem like they involve division by zero when I try to use the corollary.

$$\int_0^{2\pi} cos(cos(z))cosh(sin(z)) dz$$

The corollary that I am trying to use is:

$$f(\omega) = \frac{1}{2\pi} \int_0^{2\pi}f(\omega + \lambda e^{iz}) dz$$

For some $\lambda > 0$ and $f$ differentiable on an open set containing $\overline{D'(\omega,\lambda)}$. I think that the integral in question splits into four integrals of this form, however, two of them seem to involve division by zero when evaluating at the point needed in order to get the values. I think I must be making a mistake, if someone could show me how to use this result properly for this question that would be great.

Thanks, Lauren


This is known as Gauss's mean value theorem.

I believe there should be a contour defined in the problem of some sort. Since there isn't one listed I will use the unit circle.

Note that:

Can be turned into this:


So find the real part of the integrand and evaluate it at where the contour of the circle is centered.

Convert the integrand from trig functions into exponentials. After simplifying you will get the following expression.

After collecting the real terms one gets:

Now evaluate at the where the circle is centered, z=0 for unit disk and we get,

If you take into consideration both Real and Imaginary parts the final result of the integral is:

Also the 1/2pi factor wasn't in the original problem so the final result may be

In other words just plug in where the circle is centered into the integrand.

Note that this is my best attempt at the problem. Please let me know if there are any errors.



Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.