What is wrong in evaluating $\int_{0}^{2\pi} e^{\cos x} \cos (\sin x)\:dx$

Question

Find value of $I=\int_{0}^{2\pi} e^{\cos x} \cos (\sin x)\:dx$

I have done it as follows:

we know that $$\int_{0}^{2a}f(x) dx=2 \int_{0}^{a} f(x)dx$$ if $f(2a-x)=f(x)$

Now if $f(x)=e^{\cos x} \cos (\sin x)$, we have

$$f(2\pi-x)=f(x)$$

hence $$I=2\int_{0}^{\pi} e^{\cos x} \cos (\sin x)\:dx=2 \times Re\left\{\int_{0}^{\pi} e^{e^{ix}}dx\right\}$$

Now let

$$J=\int_{0}^{\pi} e^{e^{ix}}dx$$

Put $e^{ix}=t$, Then limits will switch over to $-1$ and $1$ Also $$e^{ix} i dx=dt$$

so

$$dx=\frac{dt}{it}$$

So

$$J=i \times \int_{-1}^{1}\frac{e^t \:dt}{t}$$

So $J$ is Purely imaginary and hence $I=0$

Can i know what went wrong in my solution?

Answer 1

Let $t(x)=e^{ix}$ for $x \in I:=[0, \pi]$.

You have assumed that $t(I)=[-1,1]$, but this is wrong.

We have $t(I)=\{z \in \mathbb C: |z|=1, Im(z) \ge 0\}$ (a "half circle").

Answer 2

The error is mentioned by @user296602 and @Fred. Below is a correct way to solve it.

Consider $$f:z\mapsto \frac{e^z}{z}$$ For $0<\epsilon<1$, we have $$0=\int_\epsilon^1\frac{e^x}{x}dx+\int_0^{\pi}\frac{e^{e^{it}}}{e^{it}}ie^{it}dt+\int_{-1}^{-\epsilon}\frac{e^x}{x}dx+\int_{\pi}^0\frac{e^{\epsilon e^{it}}}{\epsilon e^{it}}i\epsilon e^{it}dt$$ As $\epsilon\to 0$, $$\int_0^{\pi}{e^{e^{it}}}dt=\int_0^{\pi}dt+i\int_0^1\frac{e^x-e^{-x}}{x}dx$$ Thus, $$\operatorname {Re}(J)=\pi$$

Answer 3

Here is an approach that avoids complex analysis altogether.

Let $$I(\theta) = \int_0^{2\pi} e^{\theta \cos x} \cos (\theta \sin x) \, dx, \quad \theta \in \mathbb{R}$$ Observe that $I(0) = 2\pi$.

Now using Feynman's trick of differentiating the the integral sign with respect to the parameter $\theta$, we have $$I'(\theta) = \int_0^{2\pi} \frac{\partial}{\partial \theta} \left [e^{\theta \cos x} \cos (\theta \sin x) \right ] \, dx. \tag1$$ Observing that $$\frac{\partial}{\partial \theta} \left [e^{\theta \cos x} \cos (\theta \sin x) \right ] = \frac{1}{\theta} \frac{\partial}{\partial x} \left [e^{\theta \cos x} \sin (\theta \sin x) \right ],$$ (1) becomes $$I'(\theta) = \frac{1}{\theta} \int_0^{2\pi} \frac{\partial}{\partial x} \left [e^{\theta \cos x} \sin (\theta \sin x) \right ] \, dx = \frac{1}{\theta} \left [e^{\theta \cos x} \cos (\theta \sin x) \right ]_{x = 0}^{x = 2\pi} = 0.$$ Thus $I(\theta) = K$ where $K$ is a constant and can be found from $I(0) = 2\pi$. Thus $$I(\theta) = \int_0^{2\pi} e^{\theta \cos x} \cos (\theta \sin x) \, dx = 2\pi.$$ Setting $\theta = 1$ returns the integral we seek, namely $$I(1) = \int_0^{2\pi} e^{\cos x} \cos (\sin x) \, dx = 2\pi.$$