How to evaluate $\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$?

For $\alpha \in \mathbb{R}$, define $\displaystyle I(\alpha):=\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta)\; d\theta$. Calculate $I(0)$. Hence evaluate $\displaystyle\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta)\; d\theta$.

Hint: To evaluate the integral that expresses $\displaystyle\frac{dI}{d\alpha}$, consider $\displaystyle\frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta))$.

How do I do this question? I think this might have something to do with the Fundamental Theorem of Calculus, but I'm not sure.

I computed $\displaystyle I(0)=\int_{0}^{2\pi} d\theta=2 \pi$, and $\displaystyle I(1)=\int_{0}^{2\pi}e^{\cos \theta}\cos( \sin \theta) d\theta$. Following the hint I get

$$\begin{align} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) & =\alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\ & = \alpha e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + \frac{dI}{d \alpha} \cos \theta. \\ \end{align}$$

Is this correct so far?

The answers in the question referred as a duplicate does not help. I'm in a course dealing with real values, not complex.

Solution 1:

First a correction:

$$\begin{align} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) & =-\alpha \sin \theta \, e^{\alpha \cos \theta} \sin (\alpha \sin \theta) + e^{\alpha \cos \theta}\cos(\alpha \sin \theta) \alpha \cos \theta \\ \end{align}$$

Now \begin{align} \frac{dI}{d\alpha}&=\frac{d}{d\alpha}\int_{0}^{2\pi}e^{\alpha \cos \theta}\cos(\alpha \sin \theta) d\theta \\ &=\int_{0}^{2\pi}\frac{d}{d\alpha}(e^{\alpha \cos \theta}\cos(\alpha \sin \theta)) d\theta \\ &=\int_{0}^{2\pi}\cos \theta \, e^{\alpha \cos \theta}\cos(\alpha \sin \theta)- e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\sin \theta \, d\theta \\ &=\int_{0}^{2\pi}\frac{1}{\alpha} \frac{\partial}{\partial \theta}(e^{\alpha \cos \theta}\sin(\alpha \sin \theta)) d\theta \\ &=\frac{1}{\alpha} \Big[e^{\alpha \cos \theta}\sin(\alpha \sin \theta)\Big]_0^{2\pi} \\ &=0 \end{align}

So $I(\alpha)$ is actually constant.

So $I(1)=I(0)=2\pi$

So the answer is $2\pi$

Solution 2:

Alternatively, we know that $$\Re\left( e^{\Large e^{i\theta}}\right)=e^{\cos\theta}\cos(\sin\theta)$$ Using Taylor series of exponential function and Euler's formula we have $$e^{\Large e^{i\theta}}=1+(\cos\theta+i\sin\theta)+\frac{(\cos2\theta+i\sin2\theta)}{2!}+\frac{(\cos3\theta+i\sin3\theta)}{3!}+\cdots$$ Using $\displaystyle\int_0^{2\pi}\cos(n\theta)\;d\theta=0$ for $n$ is integer and $n\neq0$, we get $$\begin{align}\int_0^{2\pi}e^{\cos\theta}\cos(\sin\theta)\;d\theta&=\int_0^{2\pi}\Re\left( e^{\Large e^{i\theta}}\right)d\theta\\&=\int_0^{2\pi}\left(1 +\cos\theta+\frac{\cos2\theta}{2!}+\frac{\cos3\theta}{3!}+\cdots\right)d\theta\\&=2\pi\end{align}$$

Solution 3:

Rewrite $$ \int_0^{\large2\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\Re\left[\int_0^{\large2\pi} e^{\Large e^{i\theta}}d\theta\right]. $$ Let $$ I(\alpha)=\int_0^{\large2\pi} e^{\Large\alpha e^{i\theta}}d\theta, $$ then $$ \frac{dI}{d\alpha}=I'(\alpha)=\int_0^{\large2\pi} e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta. $$ Rewrite $$ I'(\alpha)=\frac{1}{i\alpha}\int_0^{\large2\pi} i\alpha e^{i\theta}e^{\Large\alpha e^{i\theta}}d\theta. $$ Let $x=\alpha e^{i\theta}\;\color{blue}{\Rightarrow}\;dx=i\alpha e^{i\theta}\ d\theta$, then $$ I'(\alpha)=\frac{1}{i\alpha}\left[e^{\Large\alpha e^{i\theta}}\right]_{\theta=0}^{\large2\pi}=0. $$ Thus $\Re\left[I'(\alpha)\right]=0$ and $I(\alpha)$ is a constant. Taking $\alpha=0$ yields $I(0)=2\pi$. Hence $\color{blue}{I(\alpha)=2\pi}$ and consequently $$ \int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\large\color{blue}{2\pi}. $$

Solution 4:

I simplified my approach since it was unnecessarily complicated.

As a generalization of Venus' answer , assume that $f(z)$ has a Maclaurin series expansion with real coefficients that converges absolutely on the unit circle on the complex plane. (The expansion will necessarily converge absolutely on the unit circle if it has a radius of convergence greater than $1$.)

Then $$ \begin{align} \text{Re} \int_{0}^{2 \pi} f(e^{i\theta}) \, d \theta &= \text{Re} \int_{0}^{2 \pi} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} e^{in \theta} \, d \theta \\ &= \int_{0}^{2 \pi} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \cos(n \theta) \, d \theta \\ &= \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \int_{0}^{2 \pi} \cos (n \theta) \, d \theta \\ &= f(0) \int_{0}^{2 \pi} \, d \theta + \sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} \int_{0}^{2 \pi} \cos (n \theta) \, d \theta \\ &=2 \pi f(0) + \sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} (0) \\ &= 2 \pi f(0). \end{align}$$

If we let $f(z) = e^{z}$ (a function whose Maclaurin series has an infinite radius of of convergence), we get $$ \text{Re} \int_{0}^{2 \pi} e^{e^{i \theta}} \, d \theta = \int_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) \, d \theta = 2 \pi (1) = 2 \pi.$$

Solution 5:

Here is a solution using real analysis, different from solutions already published (involving derivation under integral sign, series, etc.). This solution is based on a connection with cardinal sine function.

Let

$$f(x):=e^{-\cos x} \cos(\sin x)$$

It is periodic function with the following graphical representation:

As it is symmetrical with respect to vertical line $x=\pi$, it is sufficient to be able to compute

$$I_1:=\int_0^{\pi/2} f(x)dx \ \ \text{and} \ \ I_2:=\int_{\pi/2}^{\pi} f(x)dx$$

The result will be $2(I_1+I_2)$.

Preliminary result:

$$\int^{\color{red}{\pi/2}}_0 e^{-p\cos x} \cos(p\sin x)\,dx=\int^{\infty}_p \frac{\sin(t)}{t}dt\tag{1}$$

which is formula NT 13(26) page 486 of my 1980 edition of Gradshteyn and Ryzhik.

(where we recognize the opposite of a primitive function of cardinal sine function).

• Integral $I_1$ has the following expression by taking $p=-1$ in (1):

$$I_1=\int^{\infty}_{-1} \frac{\sin(t)}{t}dt\tag{2}$$

• It is now sufficient to establish that :

$$I_2=\int_{-\infty}^{-1} \frac{\sin(t)}{t}dt\tag{3}$$

In this way, we will have $2(I_1+I_2)=2\int_{-\infty}^{+\infty} \frac{\sin(t)}{t}dt=2 \pi$

by using the classical result $\int_{-\infty}^{\infty} \dfrac{\sin(x)}{x}dx=\pi.$

Here is the proof of (3):

$$I_2:=\int_{\pi/2}^{\pi} f(x)dx \ \ \overset{x:=\pi-X}{=} \ \ -\int_{\pi/2}^{0} e^{-\cos(X)}\cos(\sin(X))dX$$

$$I_2=\int_0^{\pi/2} e^{-\cos(X)}\cos(\sin(X))dX$$

Using (1) with $p=1$:

$$I_2=\int_1^{\infty} \frac{\sin(t)}{t}dt$$

It remains to take the change of variable $T=-t$ to obtain the desired result (3).