In textbooks on complex analysis, a typical example using Cauchy Integral Theorem is shown that
$$
\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+\theta) d \theta=0
$$
Later, I generalized it in my answer that $$\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+n\theta) d \theta=0,$$
where $r\in R^+,n\in N.$
However, I want to investigate the integral without complex analysis. So I started to try some simple cases using compound angle formula by integration by parts. $$
\begin{aligned}
I_1=\int e^{r \cos \theta} \cos (r\sin \theta+\theta) d \theta =& \int e^{r \cos \theta}[\cos (r\sin \theta) \cos \theta-\sin (r \sin \theta) \sin \theta] d \theta \\
=& \int e^{r \cos \theta} \cos (r\sin \theta) \cos \theta d \theta- \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta
\end{aligned}
$$
Fortunately,
\begin{aligned} \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta &\stackrel{IBP}{=}
-\frac{1}{r} \int \sin (r \sin \theta) d\left(e^{r \cos \theta}\right)\\&=-\frac{e^{r \cos \theta} \sin (r\sin \theta)}{r}+\int e^{r \cos \theta} \cos (r\sin \theta)\cos \theta d\theta\\
\therefore \int e^{r\cos \theta} \sin (r \sin \theta) \sin \theta d \theta&=\frac{e^{r\cos \theta} \sin (r\sin \theta)}{r}+C_1
\end{aligned}
Applying the same technique decreases $n$ to $n-1$ yields
\begin{aligned}
I_{n} &=\int e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta\\
&=\int e^{r \cos \theta}[\cos (r \sin \theta+(n-1) \theta) \cos \theta-\sin (r \sin \theta+(n-1) \theta)\sin \theta] d \theta\\
&=\int e^{r \cos \theta}[\cos (r \sin \theta+(n-1) \theta) \cos \theta d \theta-\int e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta) \sin \theta d \theta
\end{aligned}
Applying integration by parts to the last integral yields
\begin{aligned}&\int e^{r\cos \theta} \sin (r \sin \theta+(n-1) \theta) \sin \theta d \theta\\=& -\frac{1}{r} \int \sin (r \sin \theta+(n-1) \theta) d\left(e^{r \cos \theta}\right)\\=&-\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)\right] +\frac{1}{r} \int e^{r \cos \theta} \cos (r\sin \theta+(n-1) \theta)(r\cos \theta+(n-1)) d \theta \\=&-\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)\right] + \int e^{r \cos \theta} \cos (r \sin \theta+(n-1) \theta) \cos \theta d \theta+\frac{n-1}{r} I_{n-1}\end{aligned}
Putting it back yields the reduction formula
$$\boxed{I_n = \frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+(n-1) \theta)-(n-1)I_{n-1}\right]}$$
Now let’s use the reduction formula and Mathematical Induction to prove that $$J_n:=\int_0^{2\pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta=0$$
First of all, $$J_1=\left[I_1\right]_0^{2\pi}= \left[ \frac{e^{r \cos \theta} \sin (r \sin \theta)}{r}\right] _0^{2\pi}=0$$
Now assume it is true for some $k$ i.e. $J_k=0$, then
$$
J_{k+1}=\frac{1}{r}\left(\left[e^{r \cos \theta} \sin (r\sin \theta+k \theta)\right]_{0}^{2 \pi}-k J _k\right)=0
$$
Therefore it is also true for $n=k+1$ and hence by the principle of Mathematical Induction, we have $$
\boxed{\int_{0}^{2 \pi} e^{r \cos \theta} \cos (r \sin \theta+n \theta) d \theta
=0}$$
For reference, $$
\begin{aligned}
I_{2} &=\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+\theta)-I_{1}\right] \\
&=\frac{1}{r}\left[e^{r \cos \theta} \sin (r \sin \theta+\theta)-e^{r \cos \theta} \sin (r \sin \theta)\right]+C_2
\end{aligned}
$$
$$
\begin{aligned}
I_{3} &=\frac{1}{r}\left[e^{r\cos \theta}(\sin (r \sin \theta+2 \theta))-2 I_{2}\right] \\
& =\frac{1}{r^{2}}\left[r e^{r \cos \theta}(\sin (r\sin \theta+2 \theta))-2 e^{r \cos \theta } \sin (r \sin \theta+\theta)-2 e^{r \cos \theta} \sin (r\sin \theta)\right]+C_3
\end{aligned}
$$
Question: Is there an alternative proof other than MI?
Best Answer
With $$I_n(r)=\int_{0}^{2\pi}e^{r\cos\theta}\cos(n\theta+r\sin\theta)d\theta $$ it is straightforward to establish $I_{n+1}(r) =I_n’(r)$, as well as $$I_1(r)=\frac1r e^{r\cos\theta}\cos(r\sin\theta)\bigg|_{0}^{2\pi} =0$$ Thus \begin{align} I_{n+1}(r) = \frac{d^{n}}{dr^n}I_1(r)=0 \end{align}