# 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.

• Shouldn't $I(0)=\int_0^{2\pi}e^0\cos(0)d\theta$? Commented Jun 2, 2013 at 9:43
• @GitGud Is there any way I could do this question another way than the one you have answered? I haven't reached that stage in my course. Commented Jun 2, 2013 at 9:46
• you should take the hint. do the derivative they tell you to do, and see if it relates anyway to your integrand... Commented Jun 2, 2013 at 9:47
• @user4167 Have you read the other answers? Also are you sure you're supposed to do this without the tools I used? Commented Jun 2, 2013 at 9:50
• @user4167 OK. You should specify that in your question stating that the answers in the duplicate do not help, so your question doesn't get closed. Commented Jun 2, 2013 at 9:57

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 $2\pi$

• Very nice solution john. Commented Oct 24, 2016 at 6:24

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}

• Nice solution, I love it! :) Commented Nov 18, 2014 at 18:52
• Thanks @Anastasiya-Romanova. I'm glad you like it :-) Commented Nov 19, 2014 at 5:48

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 $$\int_0^{\large\pi} e^{\cos\theta}\cos(\sin\theta)\ d\theta=\large\color{blue}{2\pi}.$$

I simplified my approach.

As a generalization of Venus' answer , assume that $$\alpha$$ is a positive real number and that $$f(z)$$ is an entire function.

Then \begin{align} \int_{0}^{2 \pi} \left( f(\alpha e^{i\theta})+f(\alpha e^{-i \theta}) \right) \, \mathrm d \theta &= \int_{0}^{2 \pi} \left(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} (\alpha e^{i \theta})^{n} + \sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!} (\alpha e^{-i \theta})^{n} \right) \, \mathrm d \theta \\ &= 2\int_{0}^{2 \pi} \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \alpha^{n} \cos(n \theta) \, \mathrm d \theta \\ &= 2\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} \, \alpha^{n} \int_{0}^{2 \pi} \cos (n \theta) \, \mathrm d \theta \\ &= 2 f(0) \int_{0}^{2 \pi} \, d \theta + 2 \sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} \, \alpha^{n} \int_{0}^{2 \pi} \cos (n \theta) \, \mathrm d \theta \\ &=4 \pi f(0) + 2\sum_{n=1}^{\infty}\frac{f^{(n)}(0)}{n!} (0) \\ &= 4 \pi f(0). \end{align}

If we let $$f(z) = e^{z}$$, we get $$2 \int_{0}^{2 \pi} e^{\alpha \cos \theta} \cos (\alpha \sin \theta) \, \mathrm d \theta = 4 \pi (1) = 4 \pi.$$

And if we let $$f(z) = \cos(z)$$, for example, we get $$2 \int_{0}^{2 \pi} \cos (\alpha \cos \theta) \cosh(\alpha \sin \theta) \, \mathrm d \theta = 4 \pi (1) = 4 \pi.$$

• @FreeMind \begin{align} \sum_{k=0}^{\infty} b^{k} \cos(k \theta) &= \text{Re} \sum_{k=0}^{\infty} b^{k} e^{ik \theta} = \text{Re} \sum_{k=0}^{\infty} (be^{i \theta})^{k} \\ &= \text{Re} \ \frac{1}{1-be^{i \theta}} \\ &= \text{Re} \ \frac{1-be^{-i \theta}}{(1-be^{i \theta})(1-be^{-i \theta})} \\ &= \text{Re} \ \frac{1- b \cos \theta + i b \sin \theta}{1 - 2 b \cos \theta + b^{2}} \\ &=\frac{1- b \cos \theta}{1-2b \cos \theta + b^{2}} \end{align} Commented Dec 16, 2014 at 8:15
• @FreeMind I don't know of a list, but basically any related series can be derived by differentiating or integrating the series in my post or the series $$\sum_{k=0}^{\infty} b^{k} \sin(k \theta) = \frac{b \sin \theta}{1-2b \cos \theta + b^{2}} \ , \ |b| <1$$ with respect to $\theta$. For example, integrating the above series with respect to $\theta$ we get $$\sum_{k=1}^{\infty} \frac{b^{k} \cos (k \theta)}{k} = - \frac{1}{2} \log \left(1- 2 b \cos \theta + b^{2} \right) \ , \ |b| <1.$$ Commented Dec 17, 2014 at 6:31
• Have any idea about, math.stackexchange.com/questions/1070144/… Commented Dec 17, 2014 at 11:00
• @Machinato Alternatively, you can use the residue theorem. $$\operatorname{Re} \int_{0}^{2 \pi} f(e^{i \theta}) \, d \theta = \text{Re} \int_{|z|=1} \frac{f(z)}{iz} \, dx = \operatorname{Re} \left(2 \pi i \operatorname{Res}\left[\frac{f(z)}{iz},0 \right]\right)=2 \pi f(0)$$ I wanted to generalize the other answer without using contour integration. Commented Jun 8, 2017 at 16:43

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).

Let: $$\displaystyle \tag*{} I(n,t) = \int \limits_{0}^{2 \pi} e^{t \cos \theta} \cos( n \theta - t\sin \theta) \mathrm{d \theta}$$ $$\displaystyle \tag*{} I(n,1) = \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos( n \theta - \sin \theta) \mathrm{d \theta}$$ On differentiating both the sides, we have: $$\displaystyle \tag*{} I'(n,t) = \int \limits _{0}^{2 \pi} \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} \mathrm{d \theta}$$ We have:

$$\displaystyle \tag*{} \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} = e^{t \cos \theta} \left [(\cos (n \theta - t\sin \theta))(\cos \theta) + (\sin (n \theta - t \sin \theta))(\sin \theta)\right ]$$

Using the basic trigonometric identity, which states,

$$\displaystyle \tag*{} \cos (A-B) = \cos A \cos B + \sin A \sin B$$ We obtain: $$\displaystyle \tag*{} \dfrac{\partial( e^{t \cos \theta} \cos( n \theta - t\sin \theta))}{\partial t} = e^{t \cos \theta} \cos ((n-1) \theta - t \sin \theta)$$ Notice that: $$\displaystyle \tag{1} I'(n,t) = I(n-1,t)$$ Now, let’s derive some useful solutions $$\displaystyle \tag{2} I(0,0) = \cos(0)\int \limits _{0}^{2 \pi} \mathrm{d \theta} = 2 \pi$$ and $$\displaystyle \tag{3} I(n,0) = \int \limits _{0}^{2 \pi} \cos (n \theta) \mathrm{d \theta} = 0$$ Now, note that: $$\displaystyle \tag*{} \int \limits _{0}^{t} I'(n,t) = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$$ $$\displaystyle \tag*{} I(n,t) - \underbrace{I (n,0)}_{=0} = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$$ $$\displaystyle \tag{4} I(n,t) = \int \limits _{0}^{t} I(n-1,a) \mathrm{ da}$$ Now, to find $$I'(0,t)$$, to find this, I am going to rename the variable to apply differentiation under integral once again! Let: $$\displaystyle \tag*{} I(s) = \int \limits _{0}^{2 \pi} e^{s \cos \theta} \cos (-s \sin \theta) \mathrm{ d \theta}$$ $$\displaystyle \tag*{} I'(s) = \int \limits _{0}^{2 \pi} \dfrac{\partial (e^{s \cos \theta}\cos (-s \sin \theta))}{\partial s} \mathrm{d \theta}$$ $$\displaystyle \tag*{} \dfrac{\partial (e^{s \cos \theta}\cos (s \sin \theta))}{\partial s} \mathrm{d \theta} = e^{s \cos \theta} \cos (\theta - s \sin \theta)$$ and $$\displaystyle \tag*{} e^{s \cos \theta} \cos (\theta - s \sin \theta) \cdot s = \dfrac{\partial(e^{s \cos \theta} \sin (s \sin \theta))}{\partial \theta}$$ $$\displaystyle \tag*{} I(s) = \dfrac{1}{s} \int \limits _{0} ^{2 \pi} \partial( e^{s \cos \theta} \sin ( s \sin \theta)) = 0$$ Now, so $$I'(0,t) = 0$$ , so why not plug this into $$(4)$$, we get:

\displaystyle \tag*{} \begin{align} I(0,t)&=2\pi \\\\ I(1,t) &= 2\pi\cdot t \\\\ I(2,t) &= 2 \pi \cdot \dfrac{t^2}{2!} \\\\ I(n,t) &= 2\pi \cdot \dfrac{t^n}{t!} \end{align} and $$\displaystyle \tag*{} I(n,1) = 2 \pi \cdot \dfrac{1^n}{t!}$$ Hence, $$\displaystyle \tag*{} \boxed{\boxed{ \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos( n \theta - \sin \theta) \mathrm{d \theta} = \dfrac{2 \pi}{n!}}}$$

Now, since $$\cos x$$ is even function, we have $$\cos(-\sin x) = \cos(\sin x)$$. So we have:

$$\displaystyle \tag*{} \boxed{\boxed{ \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos(- \sin \theta) \mathrm{d \theta}=2 \pi= \int \limits_{0}^{2 \pi} e^{ \cos \theta} \cos(\sin \theta) \mathrm{d \theta}}}$$