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I'm reading the part explaining the properties of the integral of a step function in Apostol's Calculus I and he explains this property:

$$\int\limits_{ka}^{kb}s\left(\frac{x}{k}\right)dx = k\int\limits_a^bs(x)dx$$

by saying that if we distort the horizontal direction (say, the length) by a $k > 0$, it is the same as multiplying the integral by $k$. Intuitively it makes sense: if the area of a rectangle is $\text{Length} \cdot \text{Height}$, then

$$(k\cdot \text{Length}) \cdot \text{Height} = k\cdot(\text{Length} \cdot \text{Height})$$

But I have some troubles in understanding the form this property takes with ''trickier'' stretching of the interval of integration. I have been playing with the symbol $\int\limits_a^bs(x)dx$ since then, but I'm not sure whether what I did is right. For instance:

Would:

$$ \begin{align*} &1. \qquad \int\limits_{ka}^{kb}s(x)dx = k\int\limits_a^bs\left(\frac{x}{k}\right)dx\qquad \text{?}\\ &2. \qquad \int\limits_{\sqrt{a}}^{\sqrt{b}}s(x)dx = \left[\int\limits_a^bs(x^2)dx\right]^{1/2}\qquad \text{?}\\ &3. \qquad \int\limits_{a^2}^{b^2}s(x)dx = \left[\int\limits_a^bs(\sqrt{x})dx\right]^{2}\qquad \text{?}\\ &4. \qquad \int\limits_{a/k}^{b/k}s(x)dx = \frac{1}{k}\int\limits_a^bs(kx)dx\qquad \text{?} \end{align*} $$

In each case what I did was the following:

Let take $2.$ as an example. If $\sqrt{a} < x < \sqrt{b} \implies a < x^2 < b \implies x^2$ is in the domain of $s$. Then the integrand is $s(x^2)$ on $[a,b]$ and the stretching of the interval (the square root) ''drops'' to the whole integral: $\left[\int\limits_a^bs(x^2)dx\right]^{1/2}$.

If this is correct, then mechanically I know how it works but I'm not able to explain why (in particular, that part where the stretching of $[a,b]$ drops to the integral).

Thanks!!

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2 Answers 2

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None of the numbered equations ($1. - 4.$) are valid substitutions.

To avoid confusion, a different variable should be used for the substitution. In the case you present we can use the substitution $x=ku$. Then we have $$ \begin{align} \int_{ka}^{kb}s\left(\frac xk\right)\,\mathrm{d}x &=\int_{x=ka}^{x=kb}s\left(\frac xk\right)\,\mathrm{d}x\\ &=\int_{ku=ka}^{ku=kb}s\left(\frac{ku}k\right)\,\mathrm{d}ku\\ &=\int_{u=a}^{u=b}s(u)\,k\,\mathrm{d}u\\ &=k\int_a^bs(u)\,\mathrm{d}u \end{align} $$ Then, since the variable of integration is a dummy variable, we can simply replace it with any other variable, say $x$: $$ \int_{ka}^{kb}s\left(\frac xk\right)\,\mathrm{d}x =k\int_a^bs(x)\,\mathrm{d}x $$

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  • $\begingroup$ Ah, I see. I'm a little confused though: if $\mathrm{d}x$ is an infitesimal, which does not satisfy the Archimedean Property, then how is that one can carry out operations as it would ($\mathrm{d}x = \mathrm{d}ku$)? Surely I'm misunderstanding something, but I do not what. $\endgroup$
    – asd
    Feb 20, 2015 at 14:42
  • $\begingroup$ @Jazz: The change of variables in one dimension gives $\mathrm{d}u=u'\,\mathrm{d}x$ where $u=u(x)$ is a function of $x$. For example, if $u=\sin(x)$, then $\mathrm{d}u=\cos(x)\,\mathrm{d}x$. I would not think of $\mathrm{d}x$ as an infinitesimal, but as the limit of a small change in $x$. This comes from the concepts of the Riemann Integral and Riemann Sums where $$\int_a^bf(x)\,\mathrm{d}x=\sum_{k=1}^n\overbrace{f\left(a+(b-a)\frac kn\right)}^{f(x)}\overbrace{\frac{b-a}n}^{\mathrm{d}x}$$ $\endgroup$
    – robjohn ♦
    Feb 20, 2015 at 15:23
  • $\begingroup$ Sorry for my late response. Then, it is the concept of limit who allows us to apply the properties of real numbers to a differential. Is that correct? I've looked at the next pages of the chapters treating integration and there is no such a thing as Riemann Sum/Integral. In the book, integration is treated before differentiation. Maybe it has something to do with it. $\endgroup$
    – asd
    Feb 20, 2015 at 16:59
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    $\begingroup$ @Jazz: Integration before differentiation? That is really odd. You can integrate polynomials and probably a few other things, but integration by parts and substitutions are much easier to understand when you know differentiation. I'm at a loss for how to introduce integration if not as a Riemann Sum or inverse derivative. $\endgroup$
    – robjohn ♦
    Feb 20, 2015 at 17:11
  • $\begingroup$ So it seems (:. I've asked other questions similar to this one (regarding the same property) and the answers involve substitution that I do not understand yet. The book treats integration by parts and substitution techniques after treating differentiation though (:. Hopefully all of this makes sense at this stage. $\endgroup$
    – asd
    Feb 20, 2015 at 17:21
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Hint: Make the substitution $u = \frac{x}{k}$ then $du = \frac{1}{k} dx \implies kdu = dx$.

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  • $\begingroup$ By looking at later chapters in the book I saw that one can make a substitution but one needs to make some change in the dummy variable $dx$ too. I do not understand that yet /:. So far the substitution I would do with the hint would be: $\int\limits_{ka}^{kb}s(u)dx = \int\limits_a^bs(ku)dx$, but I think it's wrong. $\endgroup$
    – asd
    Feb 20, 2015 at 14:20
  • $\begingroup$ @Jazz See my edit. You're missing a $k$. Do you understand what a dummy variable is? For example $\sum_{i=0}^{\infty} i = \sum_{j=0}^{\infty} j = \sum_{k=0}^{\infty} k = \ldots$ $\endgroup$
    – Aaron Maroja
    Feb 20, 2015 at 14:31
  • $\begingroup$ Also, notice that a definite integral gives a number, then it makes no difference which variables you're using, see: $$\int_{a}^{b} f(x)dx = \int_{a}^{b} f(u) du = \ldots $$ $\endgroup$
    – Aaron Maroja
    Feb 20, 2015 at 14:38

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