# Fourier series to integral - Joseph Fourier's explanation

I do not understand the way that Joseph Fourier took from fourier series to integral

Following is from his book The Analytic Theory of Heat, 1822, translated by Freeman, 1878

Page.345, Last paragraph, Chapter IV, Section I

1. In the equation $$\frac{1}{4}\pi=\sin u + \frac{1}{3} \sin 3u + \frac{1}{5}\sin 5u + \&c.$$ we shall write instead of $u$ the quantity $\frac{x}{n}$; $x$ is a new variable, and $n$ is an infinite number equal to $\frac{1}{\mathrm{d}q}$; q is a quantity formed by the successive addition of infinitely small parts equal to $\mathrm{d}q$. We shall represent the variable number $i$ by $\frac{q}{\mathrm{d}q}$. If in the general term $\frac{1}{2i+1} \sin(2i+1)\frac{x}{n}$ we put for $i$ and $n$ their values, the term becomes $\frac{\mathrm{d}q}{2q}\sin 2qx$. Hence the sum of the series is $\frac{1}{2}\int \frac{\mathrm{d}q}{q}\sin 2qx$, the integral being taken from $q=0$ to $q=\infty$; we have therefore the equation $$\frac{1}{4}\pi=\frac{1}{2}\int_0^\infty \frac{\mathrm{d}q}{q}\sin 2qx$$ ... ...

How could Joseph Fourier make the transformation: $u \rightarrow x/n$, $i \rightarrow q/\mathrm{d}q$, $n \rightarrow 1/\mathrm{d}q$? This is Joseph's way to form his theory. Especially, I don't understand this one: "$u \rightarrow x/n$", and this line: "$n$ is an infinite number equal to $\frac{1}{\mathrm{d}q}$".

Is it a rigorous proof? Is it a very simple and natural trick for a mathematician?

When I post this, I notice the question Fourier Series to Fourier Integrals asked by Alfredoz and the answer by LVK. Actually, many text books on Fourier Analysis explain it that way, which is more plausible for me than Joseph Fourier's explanation.

I have an unusually high tolerance for such arguments because I can usually see how to use a limiting process to make sense of them. But I wouldn't want to try to salvage the argment you pointed out--it looks like more trouble than it's worth. If you're interested in a nice way to get the value of the integral, start by looking at $$F(\alpha)=\int_{-\infty}^{\infty}e^{-\alpha t}\frac{\sin t}{t}\,dt,\;\; \alpha \ge 0.$$ You can view $F(\alpha)$ as an alternating series $$F(\alpha)=\sum_{n=1}^{\infty}\int_{(n-1)\pi}^{n\pi}e^{-\alpha t}\frac{\sin t}{t}\,dt.$$ So the error caused by truncating at $k\pi$ is bounded by the next term in the series, which is an integral over $[k\pi,(k+1)\pi]$. In this way you can argue that $$\lim_{\alpha\downarrow 0}F(\alpha) = F(0).$$ And you can compute $F(\alpha)$ directly by showing that $$F'(\alpha) = -\frac{1}{1+\alpha^{2}}.$$ Integrating $F'$ gives you the value of $\pi/2$.