# Integrating $\int_0^{2\pi}e^{\cos{(x)}}\cos{(\sin{(x)})}\,dx$. Antiderivative evaluates out to be $0$ when it clearly isn't.

I have a function $$f(x)=e^{\cos{(x)}}\cos{(\sin{(x)})}$$ that I want to integrate from $$0$$ to $$2\pi$$.

Using Mathematica, the indefinite integral evaluates out to be $$\frac{-i\operatorname{Ei}(e^{ix})+i\operatorname{Ei}(e^{-ix})}{2}+C$$ where $$\operatorname{Ei}(x)$$ is the exponential integral.

When I ask Mathematica the definite integral directly, it gives a $$2\pi$$, but when I try to use the fundamental theorem of calculus, $$\implies\left.\frac{-i\operatorname{Ei}(e^{ix})+i\operatorname{Ei}(e^{-ix})}{2}\right|^{2\pi}_0=\frac{-i\operatorname{Ei}(e^{2\pi\cdot i})+i\operatorname{Ei}(e^{2\pi\cdot -i})}{2}-\frac{-i\operatorname{Ei}(e^{0i})+i\operatorname{Ei}(e^{-0i})}{2}$$ $$\implies\frac{-i\operatorname{Ei}(1)+i\operatorname{Ei}(1)}{2}-\frac{-i\operatorname{Ei}(1)+i\operatorname{Ei}(1)}{2}=0$$

Why do I get an incorrect value of 0 when I try to use the fundamental theorem of calculus on this integral?

• WolframAlpha's plot of the antiderivative suggests that it is valid only over certain intervals, one example being $(-\pi,\pi)$. Looking at the image, it can be seen that the portion of the graph over $(\pi,2\pi)$ is the negative of the portion over $(0,\pi)$, explaining why your computation evaluates to $0$. Oct 26, 2021 at 3:20
• You are likely falling victim to this problem. Long story short: when using symbolic computing software, don't use antiderivatives to compute definite integrals, unless you are absolutely certain that you are not crossing branch cuts, going around singularities, etc. Oct 26, 2021 at 4:17
• @AlannRosas But why would the correct answer be $2\pi$ if both areas cancel out? Oct 26, 2021 at 12:11
• You need to split at singularities of the expression for the antiderivative. If you call the original expression for the antiderivative $F(x)$ then you can get your desired answer as $F(2\pi^-)-F(\pi^+)+F(\pi^-)-F(0)$.
– Ian
Oct 26, 2021 at 12:30
• That doesn't really have anything to do with it being an antiderivative, it is just shorthand for $\lim_{x \to \pi^+} F(x)$.
– Ian
Oct 26, 2021 at 12:46

Graph the alleged antiderivative $$(-i\operatorname{Ei}(e^{ix})+i\operatorname{Ei}(e^{-ix}))/{2}$$:

The fundamental theorem of calculus is not applicable using this as antiderivative, since this is not a continuous function.

• Wait how does your graph demonstrate discontinuity? I think there should be a jump discontinuity at $\pi$ right? So it would be two open circles at the top and bottom and a closed circle on the x axis right? Oct 26, 2021 at 15:40
• This is how that CAS graphs a jump discontinuity. Oct 26, 2021 at 15:46
• Ah I see. That's fair. Oct 26, 2021 at 15:48

Both the comments above show why the computation failed. An alternate route is to solve it via series.

Recall that $$\sum_{k=0}^{\infty} \frac{\cos^2(\frac{kx}{2})}{k!} = \frac{e + e^{\cos(x)} \cos(\sin(x))}{2}$$

After some rearranging, integrating from $$0$$ to $$2\pi$$ yields $$(-2e \pi + \sum_{k=0}^{\infty} (\frac{2 \pi }{k!}) + ( 2 \pi +\sum_{k=1}^{\infty}\frac{\sin (2 \pi k)}{k \: k!})$$ with the $$2\pi$$ coming from the limit at $$k=0$$. The left two items cancel and the $$\sin$$ sum is just $$0$$, leaving you with $$2 \pi$$.

$$I=\int_0^{2\pi}e^{\cos{(x)}}\cos{(\sin{(x)})} dx=\Re \int_{0}^{2\pi} e^{e^{ix}}dx=\Re \left(\int_{0}^{2\pi} dx+\sum_{k=1}^{\infty} \frac{1}{k!}\int_{0}^{2\pi}e^{ikx} dx\right)=2\pi.$$ $$\int_{0}^{2\pi} e^{ikx} dx=0, k\in I$$

• I'm assuming the summation comes from the infinite series for $e^x$? Also why doesn't the fundamental theorem of calculus work on the antiderivative though? Oct 26, 2021 at 12:20
• $\int e^{e^t} dt$, is not doable. Oct 26, 2021 at 12:47
• What do you mean by doable? Oct 26, 2021 at 12:48
• it will be series and it cannot be expressed in terms of simple well known functions. $Ei(z)$ is not supposed to be a fundamental function. Bit within $[0,2\pi]$ it is doable as the answer is a well known number.: $2]pi$. Oct 26, 2021 at 12:55

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

From the Euler's formula:

$$I= \int_{0}^{2\pi} e^{\cos x}\left[\frac{e^{i\sin x}+e^{-i \sin x}}{2}\right] dx = \frac{1}{2} \int_{0}^{2\pi} \left[e^{\cos x + i \sin x} + e^{\cos x - i \sin x}\right] dx = \frac{1}{2} \int_{0}^{2\pi} \left[e^{\cos x + i \sin x} + e^{\cos(-x) + i \sin (-x)}\right] dx = \frac{1}{2} \int_{0}^{2\pi} \left[e^{e^{ix}} + e^{e^{-ix}}\right] dx$$

If we do the change of variable (see Appendix)

$$e^{ix} = z$$ $$dx = \frac{dz}{zi}$$

$$I= \frac{1}{2} \int_{0}^{2\pi} \left[e^{e^{ix}} + e^{e^{-ix}}\right] dx = \frac{1}{2i}\oint_{|z|=1} \frac{e^{z}+e^{\frac{1}{z}}}{z} dz = \underbrace{\frac{1}{2i}\oint_{|z|=1} \frac{e^{z}}{z} dz}_{A} +\underbrace{\frac{1}{2i}\oint_{|z|=1}\frac{e^{\frac{1}{z}}}{z} dz}_{B}$$

Now, apply the Cauchy's integral formula for A : $$f(a)=\frac{1}{2\pi\mathrm{i}}\oint_C \frac{f(z)}{z-a}\mathrm{d}z$$

with $$f(z) = e^{z}$$, $$a= 0$$ and $$C$$ the unit circle:

$$A = \frac{1}{2i}\oint_{|z|=1} \frac{e^{z}}{z} dz = \pi$$

For $$B$$, use the residue theorem

Expand $$\displaystyle \frac{e^{\frac{1}{z}}}{z}$$:

$$\frac{e^{\frac{1}{z}}}{z} = \frac{1}{z}+\frac{1}{z^2}+\frac{1}{z^32!}+\frac{1}{z^43!}+...$$

Therefore $$\displaystyle \operatorname{Res}\left(\frac{e^{\frac{1}{z}}}{z} ,0\right) = 1$$

Then

$$B = \frac{1}{2i}\oint_{|z|=1}\frac{e^{\frac{1}{z}}}{z} dz = \pi \operatorname{Res}\left(\frac{e^{\frac{1}{z}}}{z} ,0\right) = \pi$$ \

Hence

$$\boxed{ I = \int_{0}^{2\pi} e^{\cos x}\cos{(\sin x)} dx = 2\pi }$$

Appendix

This method will be useful in evaluating definite integrals of the type

$$\int_{0}^{2\pi} F(e^{i\theta})d\theta$$

The fact that $$\theta$$ varies from $$0$$ to $$2\pi$$ suggest that we consider $$\theta$$ as an argument of a point $$z$$ on the unit circle $$C = \left\{ z \Big| |z|=1\right\}$$ centered at the origin; hence we write $$z =e^{i\theta} \quad 0 \leq \theta \leq 2\pi$$. When we make this substitution using the equations:

$$e^{i\theta} = z$$ $$d\theta = \frac{dz}{zi}$$

The integral becomes the contour integral

$$\int_{0}^{2\pi} F(e^{i\theta})d\theta = \oint_{|z|=1} F(z)\frac{dz}{zi}$$

Conversely, if we have the complex function:

$$f(z) = \frac{F(z)}{zi}$$

and the contour $$\gamma(\theta) = e^{i\theta} \quad 0\leq \theta\leq 2\pi$$. By definiton, the contour integral in $$\gamma(\theta)$$ is:

$$\oint_{\gamma} f(z) dz = \int_{0}^{2\pi} f(\gamma(\theta))\gamma'(\theta) d\theta = \int_{0}^{2\pi} \frac{F(e^{i\theta})}{e^{i\theta}i} ie^{i\theta} d\theta = \int_{0}^{2\pi} F(e^{i\theta}) d\theta$$

So, whenever you find an integral of the type $$\int_{0}^{2\pi} F(e^{i\theta}) d\theta$$ try to evaluate it converting it in a contour integral and then applying the techniques of complex integration.

• Hm reading over this, I can't follow how you converted the regular integral to a contour integral. Would you mind expanding and elaborating that process? Oct 29, 2021 at 2:53
• Sure @Max0815, I will explain further Oct 29, 2021 at 4:09
• @Max0815 I have added an Appendix explaining the method. Additionally, a somewhat similar explanation is given in Complex Variables and Applicaitons by Churchill et. al. , third edition, page 187. Oct 29, 2021 at 5:35