# Computing $i\int\limits_0^{2\pi} sin(e^{4e^{i\theta}}) d\theta$

I wanted to calculate:

$$\int\limits_{|z|=4} \frac{\sin(e^z)}{z} dz$$

Using the Cauchy integral form, it's easy to check that:

$$=\int\limits_{|z|=4} \frac{\sin e^z}{z-(0)} = 2\pi i \sin(e^0) = 2\pi i \sin(1)$$

But... then I thought, what if I compute the contour integral directly, without the use of Cauchy's theorem?
Evaluating the parametrization $$z=4e^{i\theta}$$:

$$\begin{split} &=\int\limits_0^{2\pi} \frac{\sin\big(e^{4e^{i\theta}}\big)}{4e^{i\theta}} 4ie^{i\theta}d\theta\\ &=i\int\limits_0^{2\pi} \sin\big(e^{4e^{i\theta}}\big) d\theta \end{split}$$

and from there I'm stuck.

Any hint on the integral will be appreciated <3.

• What does $\dfrac{\sin(e^z)}z$ have to do with $\dfrac{e^z}{z+1}$? That said, I don't believe you (or anyone else) are going to evaluate that parametric integral directly. – Ted Shifrin Oct 13 '19 at 22:08
• This is the big advantage of Cauchy integral formula, it helps us with integrals which are too hard to calculate in a direct way. So why would you try other ways? – Mark Oct 13 '19 at 22:12
• Oops... that was a typo, thanks!. So, in summary, both of you think that there's no point on trying to evaluate the integral directly, right?. Thanks again for your time – holahola Oct 13 '19 at 22:41