Interchange of two improper integrals in Section 2.7. Titchmarsh's book of Riemann Zeta Function

Question

The following text is from Titchmarsh's book The Theory of the Riemann Zeta-Function:

My confusion is regarding the equality $$\int_0^{\infty} \xi^{s-1} d\xi \int_0^{\infty} f(y) \sin (\xi y) dy = \int_0^{\infty} f(y) dy \int_0^{\infty} \xi^{s-1} \sin (\xi y) d\xi.$$

Q1- Earlier in this book I encountered "since the series converges boundedly the interchange of series and integral is justified"; since, many types of interchanges I knew that would happen in later section, I studied 90 pages the entire two chapters of Ghorpade's Real Analysis. Unfortunately, the only theorem it has for interchange of two improper integrals is the condition that $f(x,y)$ being bounded and positive over the domain of integration but sine factor is not positive. My question is : is there a theorem on interchange of two Riemann improper integrals applicable to justify the exchange of integrals mentioned above? I know there are theorems using Lebesgue integral but I hope for not using Lebesgue integration, Fubini Theorem, etc. A simple readable for self-study reference in the subject helps for this specific function as well would be much appreciated.

Q2- The justification of exchange explained in the text (last paragraph in the picture) is not clear at all: For example, what is $\Delta$? Why uniform convergence is needed? Why limits $\delta \to 0$ and $\Delta \to \infty$ are taken and why they are sufficient to prove interchangeability of integrals?

To summarize: What are the theorems used in the text above?

Answer 1

1. $\delta$ and $\Delta$ are just arbitrary constants satisfying $0 < \delta < \Delta$. You may think of them as auxiliary variables that are temporarily frozen during the estimation of the integral. In the end, we will release hold of these variables and take limit as $\delta \to 0^+$ and $\Delta \to \infty$.

2. For $\xi > 0$ and for any $0 < a < b$, we have

\begin{align*} \left| \int_{a}^{b} f(y)\sin(\xi y) \, \mathrm{d}y \right| &= \left| \left[ f(y) \frac{1-\cos(\xi y)}{\xi} \right]_{y=a}^{y=b} - \int_{a}^{b} f'(y) \frac{1-\cos(\xi y)}{\xi} \, \mathrm{d}y \right| \\ &\leq \frac{1}{\xi} \left( |f(a)| + |f(b)| + \int_{a}^{b} |f'(y)| \, \mathrm{d}y \right) \\ &\leq \frac{4|f(a)|}{\xi} = \mathcal{O}\biggl(\frac{1}{a\xi}\biggr). \end{align*}

So by the Cauchy criterion, the integral $\int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y$ converges uniformly in $\xi \in [\delta, \Delta]$, with the explicit bound

$$ \left| \int_{0}^{b} f(y)\sin(\xi y) \, \mathrm{d}y - \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \right| \leq \frac{4|f(b)|}{\delta} \tag{1} $$

for any $\xi \in [\delta, \Delta]$ and $b > 0$.

3. Since $(\xi, y) \mapsto \xi^{s-1}f(y)\sin(\xi y)$ is continuous on $[\delta, \Delta] \times [0, b]$, we can interchange the order of integration to get

\begin{align*} \int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{b} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi &= \int_{0}^{b} f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y. \tag{2} \end{align*}

Now,

1. The left-hand side (LHS) of $\text{(2)}$ converges to $\int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi$ by the uniform convergence established in Item 2. (Even if you are not familiar with the uniform convergence, you can directly employ the bound $\text{(1)}$ to show that the difference tends to $0$ as $b \to \infty$.)

2. As in Item 2, we can utilize integration by parts to find that \begin{align*} \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi &= \left[ \xi^{s-1} \frac{1-\cos(\xi y)}{y} \right]_{\xi=\delta}^{\xi=\Delta} - \int_{\delta}^{\Delta} (s-1) \xi^{s-2} \frac{1-\cos(\xi y)}{y} \, \mathrm{d}\xi \\ &= \mathcal{O}(y^{-1}) \quad\text{as $y \to \infty$}. \end{align*} Using this, we find that the integrand in the right-hand side (RHS) of $\text{(2)}$ satisfies $$ f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y = \mathcal{O}(y^{-2}) \quad \text{as $y \to \infty$.} $$ In particular, the integral in RHS of $\text{(2)}$ converges as $b \to \infty$.

Combining these two observations, we get

\begin{align*} \int_{\delta}^{\Delta} \xi^{s-1} \int_{0}^{\infty} f(y)\sin(\xi y) \, \mathrm{d}y \mathrm{d}\xi &= \int_{0}^{\infty} f(y) \int_{\delta}^{\Delta} \xi^{s-1} \sin(\xi y) \, \mathrm{d}\xi \mathrm{d}y. \tag{3} \end{align*}

4. Since the LHS of $\text{(3)}$ converges to the double integral that we started from as $\delta \to 0^+$ and $\Delta \to \infty$, it suffices to prove that the RHS of $\text{(3)}$ also converges to the double integral with the interchanged order of integration. This is explained in detail in the picture, so let me skip this.