The theory of riemann zeta function titchmarsh page 15 question in the proof of the functional equation

Solution 1:

in general, for proving from scratch that you can exchange $\sum$ and $\int$ (or any two limits), start with finite bounds. clearly when $\lambda,A,B$ are finite : $$ \int_0^\lambda \sum_{n=A}^B\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=A}^B \int_0^\lambda\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=A}^B \int_0^{n\lambda} \frac{\sin(2 \pi y)}{\pi n} (y/n)^{-s-1} \frac{dy}{n} $$ $$ = \sum_{n=A}^B n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy$$

now all you need is noticing that when $Re(s) \in ]-1,0[$ : $\displaystyle\sum_{n} n^{s-1}$ is absolutely convergent, and $\displaystyle\left|\int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy\right| <C \left|\int_0^\infty\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy\right|$ (you can use integration by part for proving that the integral converges)

hence, when $Re(s) \in ]-1,0[$ the series $$\sum_{n= A}^\infty n^{s-1} \int_0^{n\lambda}\frac{\sin(2 \pi y)}{\pi } y^{-s-1}dy$$ converges absolutely (and hence it $\to 0$ when $A \to \infty$)

finally, prove that when $A \to \infty$ : $$ \int_0^\lambda \sum_{n=A}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx \to 0$$ (again you can use integration by part)

and from all this you get that :

$$ \int_0^\lambda \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \int_0^\lambda \sum_{n=1}^A\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$

$$ = \sum_{n=1}^A \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx + \int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx $$

$$= \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx - \sum_{n=A+1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx+\int_0^\lambda \sum_{n=A+1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$ $$ = \sum_{n=1}^\infty \int_0^\lambda \frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx\qquad\qquad(\text{ letting } \ A \to \infty)$$

(since the two other terms $\to 0$)

and from what Titchmarsh prove, that

$$\lim\limits_{\lambda \to \infty} \sum_{n=1}^{\infty} \frac{1}{n}\int_{\lambda}^{\infty} \frac{sin(2\pi nx)}{ x^{s+1}}dx=0$$

you get that when $Re(s) \in ]-1,0[$ :

$$ \int_0^\infty \sum_{n=1}^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx = \sum_{n=1}^\infty\int_0^\infty\frac{\sin(2 \pi n x)}{\pi n} x^{-s-1} dx$$

which (with the functional equations for the $\Gamma$ function) proves the functional equation for $\zeta(s)$.