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 $I(0)=\int_0^{2\pi}e^0\cos(0)d\theta$? – Ataraxia Jun 2 '13 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. – user4167 Jun 2 '13 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... – symplectomorphic Jun 2 '13 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? – Git Gud Jun 2 '13 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. – Git Gud Jun 2 '13 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. – juantheron Oct 24 '16 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! :) – Anastasiya-Romanova 秀 Nov 18 '14 at 18:52
• Thanks @Anastasiya-Romanova. I'm glad you like it :-) – Venus Nov 19 '14 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 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 $$\text{Re} \int_{0}^{2 \pi} e^{e^{i \theta}} \, d \theta = \int_{0}^{2 \pi} e^{\cos \theta} \cos (\sin \theta) \, d \theta = 2 \pi (1) = 2 \pi.$$

• @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} – Random Variable Dec 16 '14 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 $$\sum_{k=1}^{\infty} \frac{b^{k} \cos (k \theta)}{k} = - \frac{1}{2} \log \left(1- 2 b \cos \theta + b^{2} \right) \ , \ |b| <1.$$ – Random Variable Dec 17 '14 at 6:31
• @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. – Random Variable Jun 8 '17 at 16:43