$e, \pi, i, -1, 0$ and the super amazing 2018.

Calculus Level 4

$\large 2018 \exp \left(\int_{0}^{2 \pi} \cot (x + 2018 - 2018i) \, dx \right) =\ ?$

Notations:

$\exp (x) = e^x$ , where $e\approx 2.718$ denotes the Euler's number .
$i = \sqrt {-1}$ denotes the imaginary unit .

The answer is 2018.

1 solution

Tom Engelsman
Mar 31, 2018

Let $Z = 2018 - 2018i$ for compactness. The above definite integral evaluates to:

$\int_{0}^{2\pi} cot(x +Z) dx = ln sin(x+Z)|_0^{2\pi} = ln sin(2\pi + Z) - ln sin(Z) = ln(\frac{sin(2\pi + Z)}{sin(Z)})$ .

Completing the above expression now yields:

$2018 \cdot exp[ ln(\frac{sin(2\pi + Z)}{sin(Z)}) ] = 2018 \cdot \frac{sin(2\pi + Z)}{sin(Z)}$ ;

and since $sin(Z)$ is $2\pi$ -periodic, we conclude $sin(Z) = sin(2\pi+Z)$ and $2018 \cdot \frac{sin(2\pi + Z)}{sin(Z)} = \boxed{2018}.$

Nope, buddy. This is a wrong solution. You are missing an isolated singularity of $\cot (x)$ and the value of the integral above is $2 \pi i$ . Here you can see a right solution before applications, very interesting too

Guillermo Templado - 3 years, 2 months ago

While not technically correct, it is essentially correct, since we do have $\int \cot(x+Z)\,dx = \ln\sin(x+Z)$ on any branch of the logarithm. In particular, since $\sin(x+Z)\neq 0$ for all $x \in [0,2\pi],$ we know that $\int_0^{2\pi} \cot(x+Z)\,dx = \ln\sin(2\pi+Z) - \ln\sin(Z) + 2\pi i N = 2\pi i N$ for some $N \in \mathbb{Z}$ ( $N$ being the winding number for $\sin(x+Z)$ about $0$ for $0 \leq x \leq 2\pi$ ) Then the answer is $2018 e^{2\pi i N} = 2018$

Coincidentally, was there some vandalism on the wiki page? Most of the LaTeX functions are broken.

Edit: I don't know if the wiki page is abandoned, but I went ahead and cleaned up the LaTeX and layout so I could at least read what is there.

Brian Moehring - 2 years, 10 months ago

e , π , i , − 1 , 0 e, \pi, i, -1, 0 e , π , i , − 1 , 0 and the super amazing 2018.

The answer is 2018.

1 solution

0 pending reports

$e, \pi, i, -1, 0$ and the super amazing 2018.