Complex Functions: Calculate $\int_{0}^{2 \pi} e^{r\cos{t}}\cos{(r\sin{t}+t)} dt$

Question

Please help by telling me if my logic is correct. Thank you

Calculate the above integral. $$\int_{0}^{2\pi}e^{r \cos t}\cos{(r \sin{t}+t)}~dt$$

Note: $$\mathbb{R}(z) = r \cos{t}$$ $$\mathbb{I}(z) = r \sin{t}$$

substituting the above then $$\int_{0}^{2\pi}e^{\mathbb{R}(z)}\cos{(\mathbb{I}(z)+t)}~dt = $$ $$\int_{0}^{2\pi}e^{\mathbb{R}(z)}\left[\cos{(\mathbb{I}(z)} \cos{t}-\sin{\mathbb{I}(z)}\sin{t}\right] dt =$$ $$\int_{0}^{2\pi}e^{\mathbb{R}(z)}\left[\frac{e^{i\mathbb{I}(z)}+e^{i\mathbb{I}(z)}}{2} \cos{t}-\frac{e^{i\mathbb{I}(z)}-e^{-i\mathbb{I}(z)}}{2}\sin{t}\right]dt =$$ $$\int_{0}^{2\pi}e^{\mathbb{R}(z)}\left[\frac{e^{i\mathbb{I}(z)}+e^{i\mathbb{I}(z)}}{2} \cos{t}-\frac{e^{i\mathbb{I}(z)}-e^{-i\mathbb{I}(z)}}{2}\sin{t}\right]dt = $$ $$\frac{1}{2} \int_{0}^{2 \pi} e^{\mathbb{R}(z)}\left[e^{i\mathbb{I}(z)}\cos{t} + e^{-i\mathbb{I}(z)}\cos{t} \right] - e^{\mathbb{R}(z)}\left[e^{i\mathbb{I}(z)}\sin{t} - e^{-i\mathbb{I}(z)}\sin{t} \right] dt =$$ $$\frac{1}{2} \int_{0}^{2 \pi} \left[e^{\mathbb{R}(z)}e^{i\mathbb{I}(z)}\cos{t} + e^{\mathbb{R}(z)}e^{-i\mathbb{I}(z)}\cos{t} \right] - \left[e^{\mathbb{R}(z)}e^{i\mathbb{I}(z)}\sin{t} - e^{\mathbb{R}(z)}e^{-i\mathbb{I}(z)}\sin{t} \right] dt =$$ $$\frac{1}{2} \int_{0}^{2 \pi} e^{\mathbb{R}(z)+i\mathbb{I}(z)} \cos{t} + e^{\mathbb{R}(z)-i\mathbb{I}(z)} \cos{t}-e^{\mathbb{R}(z)+i\mathbb{I}(z)} \sin{t}+e^{\mathbb{R}(z)-i\mathbb{I}(z)} \sin{t}~dt =$$ $$\frac{1}{2} \int_{0}^{2 \pi} e^{z} \cos{t} + e^{\bar{z}} \cos{t}-e^{z} \sin{t}+e^{\bar{z}} \sin{t}~dt =$$ $$\frac{1}{2} \int_{0}^{2 \pi} e^{z}\left( \cos{t}-\sin{t} \right) + e^{\bar{z}}\left( \cos{t}+\sin{t} \right) dt =$$ $$\frac{1}{2} \int_{0}^{2 \pi} e^{re^{it}}\left( \cos{t}-\sin{t} \right) + e^{re^{-it}}\left( \cos{t}+\sin{t} \right) dt =$$ now...is it right to say that because $\cos~\&~\sin$ are both bound above by 1. we can say that $$\frac{1}{2} \int_{0}^{2 \pi} e^{re^{it}}\left( \cos{t}-\sin{t} \right) + e^{re^{-it}}\left( \cos{t}+\sin{t} \right) dt \leq \frac{1}{2} \int_{0}^{2\pi} 2e^{re^{it}}dt = 0$$

via the Cauchy's theorem.

Answer 1

You can't calculate the value if the integral by just finding its bound. You need his friend: $$\int_0^{2\pi} e^{r \cos t} \cos(r\sin t + t) dt + i\int_0^{2\pi}e^{r \cos t} \sin(r \sin t + t) dt = \int_0^{2\pi} e^{r\cos t + i r\sin t + it} dt= \int_0^{2\pi} e^{re^{it}}e^{it}dt$$

This integral is zero. Can you take it from here?