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Let $[a, b]$ be a closed interval of the real line and let a sequence as $$a = x_0 \le t_1 \le x_1 \le t_2 \le x_2 \le \cdots \le x_{n-1} \le t_n \le x_n = b . \,\!$$ This partitions the interval $[a, b]$ into $n$ sub-intervals $[x_{i−1}, x_i]$ indexed by $i$; moreover, we take $t_i \in (x_{i−1}, x_i)$. Let $$\sum_{i=0}^{n} f(t_i) \Delta_i $$ where $\Delta_i=x_i-x_{i-1}$; if $x_i=a+i h$ and $h=\frac{b-a}{n}$ we define $$S_n = h \sum_{i=0}^{n} f(t_i)\ \ \ \ (*)$$ If $f$ is continue on $[a,b]$ then $$\lim_{n\rightarrow \infty} S_n=\int_a^b f(x) dx\ \ \ \ (**)$$ I would like to get (*) and (**) for $a=-\infty$, $b=+\infty$ where $f$ continue and $\in L^1(\mathbb R)$. Any suggestions please?

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For the case that you want $\int_{a}^{\infty} f(x)dx$ the usual way to do this is to compute $\lim_{B\to \infty} \int_{a}^{B} f(x)dx$ then apply your formula for the finite integral $\int_{a}^{B}f(x)dx$. You end up with a double limit that looks like:

$$\lim_{B\to\infty} \lim_{n\to\infty} S_{n}.$$

If you want infinite limits on both ends, then break up the intervals into two pieces $$\lim_{A\to-\infty} \int_{A}^{0} f(x)dx + \lim_{B\to\infty}\int_{0}^{B}f(x)dx.$$

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You could add the terms until you reach a point that you are satisfied. Or you can use convergent series approximation techniques.

https://en.wikipedia.org/wiki/Shanks_transformation

https://en.wikipedia.org/wiki/Richardson_extrapolation

https://en.wikipedia.org/wiki/Series_acceleration

You can research more about these.

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