I’m finding it hard to convert a limit of sum which involves a composite function into a definite integral. For normal functions we can make use of this definition $$ \lim_{n\to \infty} \frac{b-a}{n} \sum_{i=1}^{n} f \left( a+ \left( \frac{b-a}{n} \right) i \right)= \int_{a}^{b} f(x) dx$$
But let’s consider I have something like this: $$\lim_{n\to \infty} \frac{x}{n}\left[ \ln \Bigg(1 + \dfrac{x^2}{n^2} \Bigg) + \ln\Bigg(1+ \dfrac{4x^2}{n^2}\Bigg) + \ln\Bigg(1+ \dfrac{9x^2}{n^2}\Bigg) ... \ln\Bigg(1+ \dfrac{n^2x^2}{n^2}\Bigg) \right]$$ Well, the above thing is not of the standard form, because it can be written like $$ \lim_{n\to \infty} \frac{x}{n} \sum_{i=1}^{n} \ln \left( 1+ \left(\frac{ix}{n} \right) ^2 \right) $$ Where, in my opinion, $x/n$ is acting as $\frac{b-a}{n}$ but the problem is inside the argument we have $\left(\frac{b-a}{n} i \right)^2$ but according to the standard form we ought to have just $\frac{b-a}{n} i $.
I asked this to someone he said the above said limit can be written as the integral $$\int_{0}^{x} \ln(1 +t^2) dt$$ And I really don’t know how and it’s my question to know how.
There is one another example, how can I write $$\int_{0}^{x} \sin (t^2) dt$$ as limit of a sum? The most I can think is $$ \lim_{n\to \infty} \frac{x}{n} \left[ \sin (x/n) + \sin(2x/n) .... + \sin(nx/n) \right]$$ but I know it’s equal to $$\int_{0}^{x} \sin (t) dt $$ Please help.
The simplest form of a riemann sum is $$\lim_{n \to \infty} \frac1n \sum_{r=1}^n f\left(\frac rn\right) = \int_0^1 f(u) du$$
Looking at the first example you gave, you can treat $x$ as a constant here. $$\lim_{n\to \infty} \frac{x}{n} \sum_{r=1}^{n} \ln \left(1+ \left(\frac{r}{n} \right)^2 x^2\right)$$
here, $f(u) = \ln( 1 + u^2x^2)$. This expression becomes $$x\int_0^1 \ln(1+u^2x^2) du$$ If we let $ux = t \implies xdu = dt$, then the integral becomes $$\int_0^x \ln(1+t^2)dt$$
Taking the $\sin(t^2)$ example, to write it as a limit of a sum, you can try $$\lim_{n \to \infty} \frac{x}{n} \sum_{r=1}^\infty \sin\left(\left(\frac rn \right)^2 x^2 \right)$$ This simplifies to $$x\int_0^1\sin(u^2 x^2)dx = \int_0^x \sin(t^2) dt$$ using the same substitution we made in the previous example.