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Improper integral $\int_0^\infty \frac{\sin(x)}{x}dx$ – Showing convergence.
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Currently, I am taking a course where we defined the multidimensional Riemann Integral of a map $f:\mathbb{R}^n \to \mathbb{R}$ as the limit

$$\lim_{\varepsilon \to 0}\phantom{a}\varepsilon^n \sum_{x \in \mathbb{Z}^n}f(\varepsilon x)$$

If we require that $f$ is continuous and has a compact support, it is guaranteed that the above limit exists.

Unfortunately, I do not know any references where the multidimensional integral is defined in this particular way. My Question:

Does anyone knows a reference (e.g. a textbook or available lecture notes) where the above definition of the multidimensional integral is discussed in some detail?

- Showing that $\int\limits_{-a}^a \frac{f(x)}{1+e^{x}} \mathrm dx = \int\limits_0^a f(x) \mathrm dx$, when $f$ is even
- Evaluating some integrals
- The product of two Riemann integrable functions is integrable
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- An Integral involving $e^{ax} +1$ and $e^{bx} + 1$
- Compute $\lim_{n\to\infty} \int_{0}^{\pi/4}\tan^n x \ dx$

- Relationship among the function spaces $C_c^\infty(\Omega)$, $C_c^\infty(\overline{\Omega})$ and $C_c^\infty(\Bbb{R}^d)$
- Find $\frac{d}{dx}\left(f_1(x)f_2(x)\cdots f_n(x)\right)$.
- Do all polynomials with order $> 1$ go to $\pm$ infinity?
- Can a function that has uncountable many points of discontinuity be integrable?
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- Evaluate $\int \cos(3x) \sin(2x) \, dx$.
- Concerning Carathéodory's criteria of differentiability and a proof that differentiable implies continuous
- Express $C$ in terms of the sets $A_n$
- lim inf $|a_n|=0 \implies \sum_{k=1}^\infty a_{n_k}$ converges
- Volume of $T_n=\{x_i\ge0:x_1+\cdots+x_n\le1\}$

I can’t give you a reference on it, but I can offer some explanation. First of all, as a “definition” of multivariate integration, it is clearly inadequate. It depends heavily on continuity, somewhat less heavily on compact support, and it only defines integration over all of $\Bbb R^n$. However, given that $f$ is continuous with compact support, it is a perfectly acceptable means of calculating the integral of $f$ over $\Bbb R^n$.

Second, to call it the “Riemann” integral is definitely inaccurate. Riemann specifically and intentionally defined his integral to be more general than this. A better name for this is the Cauchy integral, as it represents only a minor generalization of Cauchy’s original definition of integration (since Cauchy was dividing up a finite interval into an integral number of subintervals of equal size, his values of $\epsilon$ were restricted to the length of the interval divided by some positive integer). The problem with the Cauchy integral was that it allowed pathological functions to “exploit” its summation strategy to produce undesirable results. Therefore to prove almost anything useful, you had to assume that $f$ was continuous. But since many useful functions are not continuous, this was a problem. That is why Riemann developed his version of the integral. The more general summing strategy did not allow the truly pathological functions to be integrable, allowing looser conditions on theorems.

So what is going on here? Look first at the one dimensional case:

$$\begin{align}\int_{-\infty}^\infty f(x)\ dx &= \lim\limits_{\epsilon \to 0}\;\;\epsilon \sum\limits_{k=-\infty}^\infty f(k\epsilon)\\&=\lim\limits_{\epsilon \to 0}\; \sum\limits_{k=-\infty}^\infty f(k\epsilon)\epsilon\\&=\lim\limits_{\epsilon \to 0}\; \sum\limits_{k=-\infty}^\infty f(k\epsilon)\left((k+1)\epsilon – k\epsilon\right)\end{align}$$

where I hope you can spot the usual form for an integral summation. Note that since $f$ has compact support, the summation is actually finite, as for sufficiently high or low $k,\ f(k\epsilon) = 0$. Since $f$ is continuous, it does not matter where in the interval $f$ is evaluated, so this expression just uses the value at the lower end.

For two dimensions, the sum divides the plane into squares of sidelength $\epsilon$, and multiplies the height of the function at the lower left corner corner of the square by the area of the square to approximate the volume under the function. Letting the sidelengths go to $0$, this converges to the integral:

$$\begin{align}\iint f\ dA

&= \lim\limits_{\epsilon \to 0} \sum\limits_{i=-\infty}^{\infty}\sum\limits_{j=-\infty}^{\infty} f(i\epsilon, j\epsilon)\epsilon\cdot\epsilon\\

&= \lim\limits_{\epsilon \to 0} \sum\limits_{(i,j) \in \Bbb Z^2} f(i\epsilon, j\epsilon)\epsilon^2\\

&= \lim\limits_{\epsilon \to 0}\ \epsilon^2\sum\limits_{(i,j) \in \Bbb Z^2} f(\epsilon(i,j))\\

&= \lim\limits_{\epsilon \to 0}\ \epsilon^2\sum\limits_{x \in \Bbb Z^2} f(\epsilon x)

\end{align}$$

Higher dimensions are the same.

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