Calculate :$\int_{0}^{2\pi }e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$

Question

Calculate: $$\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)\mathrm{d}t$$

My try:

$\displaystyle\int_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\cos(\mathfrak{I\textrm{z}}+3(-i\log z)dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}}^{(\mathfrak{I\textrm{z}}+3(-i\log z))i}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}}z^{3}dz\\ \displaystyle \int_{|z|=R}e^{\mathfrak{R\textrm{z}}}\mathfrak{R\textrm{e}^{\mathfrak{I\textrm{z}}}R}z^{3}dz$

and there is nothing here that is not holomorphic, therefore according to Cauchy theorem it must be exactly $0$.

Answer 1

The idea is right here but the proof steps could use a little polishing. More specifically, note that the integral is connected to a nicer looking form:

$$\int_{0}^{2\pi}e^{R\cos t}\cos(R\sin t+3t)dt=\Re\int_{0}^{2\pi}e^{R\cos t+iR\sin t+3it}dt=\Re\int_{0}^{2\pi}e^{Re^{i t}}e^{i3t}dt$$

This can be transformed to a contour integral on the unit circle via the substitution $z=e^{it}$, under which we conclude that

$$\int_{0}^{2\pi}e^{Re^{it}}e^{i3t}dt=\frac{1}{3i}\oint_{|z|=1}z^3e^{Rz}dz=0$$

and indeed, the integral presented above is identically zero.

Answer 2

Let $I = \intop_{0}^{2\pi}e^{R{ {\cos t}}}\cos(R\sin t+3t)dt$ and $J = \intop_{0}^{2\pi}e^{R{ {\cos t}}}\sin(R\sin t+3t)dt$. Then $$I+iJ = \int_0^{2\pi} e^{R\cos t}e^{i(R\sin t + 3t)} dt = \int_0^{2\pi} e^{Re^{it}}e^{3it} dt = \frac{1}{i} \int_{|z|=1} e^{Rz}z^2 dz = 0.$$ So $I = 0$ (and also $J =0$).

Answer 3

Some progress:

Let $$f(t)=e^{r\cos t} \cos(r \sin t+3t)$$ $$f(2\pi-t)=e^{r \cos t} \cos[-r\sin t+6\pi-3t]=f(t)$$ According to the property $$\int_{0}^{2a} f(x) dx=\int_{0}^{a}[f(x)+f(2a-x] dx~~~~~(1)$$ So $$I=\int_{0}^{2\pi} f(t) dt=2\int_{0}^{\pi} f(t) dt$$ $$f(\pi-t)=e^{-r\cos t} \cos[r\sin t+3(\pi-t)]=-e^{-r\cos t} \cos[r\sin t-3t)]$$ $$\implies f(\pi-t)=-e^{-r\cos t}[ \cos(r\sin t) \cos 3t+\sin (r\sin t)\sin 3t]$$ $$f(t)=e^{r\cos t}[[ \cos(r\sin t) \cos 3t-\sin (r\sin t)\sin 3t]$$ $$g(t)=f(t)+f(\pi-t)=2\cos(r \sin t) \cos 3t \sinh(r\cos t)-2\sin(r\sin t)\sin 3t \cosh (r \cos t)$$ Next, use (1) Then $$I=2\int_{0}^{\pi} f(t) dt=2\int_{0}^{\pi/2} [f(t)+f(\pi-t)]dt=2\int_{0}^{\pi/2} g[t] dt$$

The answer need to be zero, I don't know how. If some one can take it from here.

Answer 4

To add one more solution, I want to generalize the result $$ I_n:=\int_{0}^{2 \pi} e^{R \cos t} \cos (R \sin t+n t) d t=0 $$

Proof: Before doing the evaluation, we need its partner integral

$$ J_n:=\int_{0}^{2 \pi} e^{R \cos t} \sin (R \sin t+n t) d t$$ Combining them together yields $$ \begin{aligned} I_n+iJ_n=& \int_{0}^{2 \pi} e^{R \cos t} \cos (R \sin t+n t) d t+i \int_{0}^{2 \pi} e^{R \cos t} \sin (R \sin t+n t) d t \\ =& \int_{0}^{2 \pi} e^{R \cos t} e^{(R \sin t+n t) i} d t \\ =& \int_{0}^{2 \pi} e^{R e^{i t}} \cdot e^{n t i} d t \\ =& \oint_{|z|=1}^{} e^{ R z}z^n d z \quad \text { where } z=e^{i t} \\ \end{aligned} $$

By the Cauchy Integral Theorem , we have $$I_n+iJ_n=0 \Rightarrow I_n=J_n=0 $$

Hence $$\int_{0}^{2 \pi} e^{R \cos t} \cos (R \sin t+n t) d t=0.$$ By the way, $$\int_{0}^{2 \pi} e^{R \cos t} \sin (R \sin t+n t) d t=0$$