Sia \( f: ]0,\pi] \rightarrow \mathbb{R} \) la funzione definita da \( f(t) = 0 \) se \( t=0 \) e \( f(t) = \frac{1}{2\sin(\frac{t}{2})} - \frac{1}{t} \) se \( 0<t \leq \pi \)
Dimostra che \[ \text{1)}\ \lim\limits_{n\to \infty} \int_{0}^{\pi} f(t) \sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}dt = 0 \]
\[ \text{2)}\ \lim\limits_{n\to \infty} \int_{0}^{(n+1/2)\pi}\frac{\sin t}{t}dt = \frac{\pi}{2} \]
\[ \text{3)}\ \int_{0}^{\infty}\frac{\sin t}{t}dt = \int_{0}^{\infty}\frac{\sin^2 t}{t^2}dt =\frac{\pi}{2} \]
Per il primo io ho pensato a questo:
\[\lim\limits_{n\to \infty} \int_{0}^{\pi} f(t) \sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}dt = \lim\limits_{n\to \infty} \int_{0}^{\pi} \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{2\sin (\frac{t}{2})} - \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{t}dt\]
Spezziamo l'nitegrale e studiamo il primo integrale, sia inoltre \( \varepsilon \in ]0,\pi] \)
\[ \lim\limits_{\varepsilon \to 0} \lim\limits_{n \to \infty} \int_{\varepsilon}^{\pi} \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{2\sin (\frac{t}{2})} dt \], Abbiamo inoltre la seguente eguaglianza con \( t \in ]0, \pi] \) e \( n >0 \)
\[ \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{2\sin (\frac{t}{2})} = \frac{1}{2} + \sum\limits_{k=1}^{n} \cos(kt) \Leftrightarrow \sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix} = \sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix} + \sum\limits_{k=1}^{n} 2\sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix}\cos(kt)\]
\[ = \sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix} + \sum\limits_{k=1}^{n} \sin \begin{pmatrix}
\frac{t}{2} + kt
\end{pmatrix}+\sin \begin{pmatrix}
\frac{t}{2} - kt
\end{pmatrix}= \sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix} + \sum\limits_{k=1}^{n} \sin \begin{pmatrix}
\frac{t}{2} + kt
\end{pmatrix}-\sin \begin{pmatrix}
kt-\frac{t}{2}
\end{pmatrix}\]
\[=\sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix}+\sin \begin{pmatrix}
\frac{3t}{2}
\end{pmatrix}-\sin \begin{pmatrix}
\frac{t}{2}
\end{pmatrix}+\ldots + \sin \begin{pmatrix}
\frac{(2n-1)t}{2}
\end{pmatrix}+\sin \begin{pmatrix}
\frac{(2n+1)t}{2}
\end{pmatrix}-\sin \begin{pmatrix}
\frac{(2n-1)t}{2}
\end{pmatrix}\]
Dunque tornando all'integrale
\[ \lim\limits_{\varepsilon \to 0} \lim\limits_{n \to \infty} \int_{\varepsilon}^{\pi} \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{2\sin (\frac{t}{2})} dt=\lim\limits_{\varepsilon \to 0} \lim\limits_{n \to \infty} \frac{t}{2} + \int_{\varepsilon}^{\pi} \sum\limits_{k=1}^{n} \cos(kt) dt\]
\[=\lim\limits_{\varepsilon \to 0} \lim\limits_{n \to \infty} \frac{\pi}{2} - \frac{\varepsilon}{2} + \sum\limits_{k=1}^{n} \frac{\sin(k\pi)}{k} - \frac{\sin(k\varepsilon)}{k} = \frac{\pi}{2}\]
Pertanto dobbiamo avere che \[ \lim\limits_{\varepsilon \to 0} \lim\limits_{n \to \infty} \int_{\varepsilon}^{\pi} \frac{\sin\begin{bmatrix}
\begin{pmatrix}
n+ \frac{1}{2}
\end{pmatrix}t
\end{bmatrix}}{t} dt=\frac{\pi}{2} \]
Ma non riesco a dimostrarlo... suggerimenti? Mi sembra anche sensato in quanto l'esercizio 2) è praticamente identico. Qui le altre domande, il fatto che l'integrale qui sopra faccia \( \frac{\pi}{2} \) implica il 2)? E il 2) implica direttamente che \[ \int_{0}^{\infty} \frac{\sin t}{t} dt = \frac{\pi}{2} \] oppure no??