Hello maths community!  One day I was solving a geometry problem and I thought I had found a way of solving it. When I was solving the problem, I kind of invented a new way of finding an area of a shape that is related to circle but just recently I realized that I was wrong. The method I invented was invalid but I just don't understand why wouldn't it work... I won't describe the problem here but I am going to show an example of the method I invented.  Warning. The following methods are invalid!  To find the area of a circle with my method is to first split the circle in half and first find the area of that and then multiply it by 2 in the end to make our life easier.  A half circle is made out of bunch of vertical "lines" and I thought one could use integrals to add these "lines" up to get the area. To do this we need to find a function to get the length of one "line" and then integrate the function. If we draw a right triangle inside the half circle, the task will become easy.      Now, to get $l$, we use sine: $l=r\cdot \sin(φ)$  Then we take the integral with respect to $φ$. I guess this part is the one causing problems...  $$\int_{0}^\pi l \, dφ=\int_0^\pi r\cdot \sin(φ) \, dφ=r\int_0^\pi \sin(φ) \, dφ = r (-\cos(\pi) +\cos(0))=2r$$  Well we know already that the area of a half circle is definitely $2r$  and if we multiplied it by $2$ we would get $4r$ and $4r \neq \pi r^2$  With my logic this would seem very valid and I would just like to know why this doesn't work. Thanks!
I was looking for proofs using Calculus for the area of a circle and come across this one $$\int 2 \pi r \, dr = 2\pi \frac {r^2}{2} = \pi r^2$$  and it struck me as being particularly easy. The only other proof I've seen was by a teacher and it involved integrating $x = \sqrt{r^2 - y^2}$ from $-1$ to $1$, using trig substitutions and then doubling the area to get $\pi r^2$ but the above proof seemed much more straight forward.  Is it a valid proof, or is it based on circular logic or some other kind of fallacy?
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As stated at the top I'm looking for a solution in python in QGIS. $\sum_{n=-\infty}^{\infty}\frac{1}{(u +n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$ I've already see a  by Marko Riedel which I list it follows:     The standard way to treat these sums is to integrate   $$ f(z) = \frac{1}{(z+\alpha)^2} \pi \cot(\pi z)$$   along a contour consisting of a circle of radius $R$ and with $R$ going to infinity and hence being larger than $\alpha$, where the circle does not pass through the poles on the real axis.   Now along the semicircle in the upper half plane we have   $$|f(z)| \le \frac{1}{(R-|\alpha|)^2}\pi \left|\frac{e^{i\pi R\exp(i\theta)} + e^{-i\pi R\exp(i\theta)}} {e^{i\pi R\exp(i\theta)} - e^{-i\pi R\exp(i\theta)}}\right|= \frac{1}{(R-|\alpha|)^2} \pi  \left|\frac{e^{2i\pi R\exp(i\theta)}+1}{e^{2i\pi R\exp(i\theta)}-1}\right| &lt; \frac{1}{(R-|\alpha|)^2} \pi   \frac{1+e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}} {1-e^{-2\pi R\sin(\theta)}e^{2i\pi R\cos(\theta)}}$$   This last term is clearly $O(1/R^2)$ as $R$ goes to infinity as the quotient of the two exponentials goes to one since $\exp(-R)$ vanishes and there is no singularity when $\theta = 0$ or $\theta = \pi$ as $R\cos\theta = \pm R$, which is not an integer by the assumption that the circle avoids the poles and hence cannot be one.     My question: It is clear that for each $z=Re^{i \theta}$ , $f(z)=O(1/R^2)$ ，but can this guarantee that the last term on RHS is uniform bounded ? 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