I'm stuck when I try to solve this integral:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx$$
I try this:
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx=\lim_{n\rightarrow\infty}\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}\frac{1}{n}=\lim_{n\rightarrow\infty}\frac{1}{n}\sum_{i=1}^{n}\frac{1}{1+(\frac{i}{n})^{2}}$$
Can someone help with the next step? I'm stuck.
We can proceed as follows. First, we expand the term $\frac{n}{k^2+n^2}$ in the geometric series
$$\frac{n}{k^2+n^2}=\sum_{\ell =0}^N (-1)^\ell \frac{k^{2\ell}}{n^{2\ell +1}}+(-1)^{N+1}\frac{k^{2N+2}}{n^{2N+1}(k^2+n^2)}\tag 1$$
Then, summing $(1)$ reveals
$$\begin{align} \sum_{k=1}^n \frac{n}{k^2+n^2}&=\sum_{k=1}^n\sum_{\ell =0}^N (-1)^\ell \frac{k^{2\ell}}{n^{2\ell +1}}+\frac{(-1)^{N+1}}{n^{2N+1}}\sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\\\\ &=\sum_{\ell =0}^N \frac{(-1)^\ell}{n^{2\ell +1}} \sum_{k=1}^n k^{2\ell}+\frac{(-1)^{N+1}}{n^{2N+1}}\sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\tag2 \end{align}$$
Note that we have the relationships
$$\sum_{k=1}^n k^{2\ell}=\frac{n^{2\ell +1}}{2\ell +1}+O(n^{2\ell}) \tag3$$
and
$$\frac{n^{2N+1}}{2(2N+3)}\le \sum_{k=1}^n\frac{k^{2N+2}}{k^2+n^2}\le \frac{n^{2N+1}}{2N+3}\tag 4$$
Using $(3)$ and $(4)$ in $(2)$ and letting $N\to \infty$ yields
$$\sum_{k=1}^n \frac{n}{k^2+n^2}=\sum_{\ell =0}^\infty \frac{(-1)^\ell}{2\ell +1}+O\left(\frac1n\right)$$
whereupon letting $n\to \infty$ we obtain
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{n}{k^2+n^2}=\sum_{\ell =0}^\infty \frac{(-1)^\ell}{2\ell +1}$$
Finally, recalling that the Taylor series for $\arctan(x)=\sum_{\ell=0}^\infty \frac{(-1)^\ell\,x^{2\ell +1}}{2\ell +1}$, we see that
$$\lim_{n\to \infty}\sum_{k=1}^n \frac{n}{k^2+n^2}=\pi/4$$
as expected!
Please, "Riemann sum" does not mean that you have to use a partition of the interval into equal parts! There's a lot of freedom: a Riemann sum for $\int^b_a f(x)\,dx$ is any sum of the form $$R_n=\sum^{n-1}_{k=0}f(\xi_k)\,(x_{k+1}-x_k),$$ where $x_0=a$ and $x_n=b$ and $\xi_k\in[x_k,x_{k+1}]$ (and $x_k<x_{k+1}$, naturally). Who stops us to choose $x_k=\tan\phi_k$, with $\phi_k=\frac{k\,\pi}{4\,n}$? Then, $$x_{k+1}-x_k=\tan(\phi_{k+1}-\phi_k)\,(1+\tan\phi_k\,\tan\phi_{k+1}),$$ and if we choose $\xi_k$ so that $\xi^2_k=\tan\phi_k\,\tan\phi_{k+1}$ (that's between $x_k$ and $x_{k+1}$ because of the monotony and continuity of $\tan$ in $[0,\pi/4]$), our Riemann sum becomes $R_n=n\,\tan\frac{\pi}{4\,n}$, and clearly $R_n\rightarrow\pi/4$ as $n\rightarrow\infty$ (the word "clearly" masks the fact that even that last step is more complex than to calculate the integral via the Fundamental Theorem, but don't blame me, I didn't invent that exercise!).
We want
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx$$
Let $x=\tan \theta,\quad dx=d\theta/\cos^2\theta$, then
$$\int_{0}^{1}\frac{1}{1+x^{2}}dx=\int_0^{\pi/4}\frac{d\theta}{(1+\tan^2\theta)\cos^2\theta}=\int_0^{\pi/4}d\theta=\frac{\pi}{4}$$
In fact,
$$\int_{0}^{a}\frac{1}{1+x^{2}}dx=\tan^{-1}a$$ $$\int_{0}^{\infty}\frac{1}{1+x^{2}}dx=\frac{\pi}{2};\quad \int_{0}^{\sqrt{3}}\frac{1}{1+x^{2}}dx=\frac{\pi}{3};\quad \text{etc.}$$
Not sure why you can't use the well-known antiderivative $arctanx$, to get $$\int_0^1\frac1 {1+x^2}dx=[arctanx]_0^1=\frac \pi4-0=\frac \pi4$$...