I want to compute an integral, but I don't know how. The exercice is about the proof of this equality : $ \int _0 ^ \pi \frac {\sin(t) }{t} dt = \frac{\pi}{2} $. One step require to show this equality :
$$ \lim_{n \rightarrow \infty} \int _0 ^ \pi \frac {\sin((n+1/2)t) }{t} dt = \frac{\pi}{2} $$
Do you have any idea? I tried to use an integration by part, ive found :
$$ -[ \frac{cos((n+1/2)t)}{(n+1/2)t} ]^\pi_0 - \int _0 ^ \pi \frac {\cos((n+1/2)t) }{(n+1/2) t^2} dt $$
but this one is not defined... wheras the first expression is defined (because in 0 $\ sin(t) / t $ is defined.
I don't know what to do.
Thank you :)
Start by considering the rather unusual function $$f(t) = \frac{1}{t} - \frac{1}{2 \sin (t/2)}, \quad 0 < t \leqslant \pi.$$ At $t = 0$ the value for the function is understood as its limiting value as $t \to 0^+$ so in this case we may write $f(0) = \lim_{t \to 0^+} f(t) = 0$. Thus $f$ will be continuous for all $t \in [0,\pi]$ and its derivative can be shown to also be continuous for all $t$ on the interval $[0,\pi]$.
Now consider the integral $$\int_0^\pi \sin \left (\left (n + \frac{1}{2} \right ) t \right ) f(t) \, dt.$$
Integrating by parts we have \begin{align*} \int_0^\pi \sin \left (\left (n + \frac{1}{2} \right ) t \right ) f(t) \, dt &= \left [\frac{-1}{n + \frac{1}{2}} \cos \left ( \left (n + \frac{1}{2} \right ) t \right ) \cdot f(t) \right ]_0^\pi\\ & \qquad + \frac{1}{n + \frac{1}{2}} \int_0^\pi \cos \left ( \left (n + \frac{1}{2} \right ) t \right ) \cdot f'(t) \, dt. \end{align*} As $f(0) = 0$ and $\cos (n + \frac{1}{2}) \pi = 0$, the above integral reduces to $$\int_0^\pi \sin \left (\left (n + \frac{1}{2} \right ) t \right ) f(t) \, dt = \frac{1}{n + \frac{1}{2}} \int_0^\pi \cos \left ( \left (n + \frac{1}{2} \right ) t \right ) \cdot f'(t) \, dt. \tag1$$
Now as $f'(t)$ is continuous on $[0,\pi]$, $|f'(t)|$ will be bounded by some constant $M > 0$. Thus $|f'(t)| \leqslant M$ and we have \begin{align*} \left |\int_0^\pi \cos \left (\left (n + \frac{1}{2} \right ) t \right ) f'(t) \, dt \right | &\leqslant \int_0^\pi \left |\cos \left (\left (n + \frac{1}{2} \right ) t \right ) \right | \cdot |f'(t)| \, dt\\ &\leqslant \int_0^\pi |f'(t)| \, dt, \quad \text{since} \,\, |\cos (n + \frac{1}{2})t| \leqslant 1\\ &\leqslant M \int_0^\pi dt, \quad \text{since} \,\, |f'(t)| \leqslant M\\ &= M \pi, \end{align*} or $$\left |\frac{1}{n + \frac{1}{2}} \int_0^\pi \cos \left (\left (n + \frac{1}{2} \right ) t \right ) f'(t) \, dt \right | \leqslant \frac{M \pi}{n + \frac{1}{2}},\tag2$$ after dividing both side by the positive term $n + \frac{1}{2}$.
In the limit as $n \to \infty$ from (2) we see that $$\left |\frac{1}{n + \frac{1}{2}} \int_0^\pi \cos \left (\left (n + \frac{1}{2} \right ) t \right ) f'(t) \, dt \right | \to 0,$$ and so from (1) conclude that $$\lim_{n \to \infty} \int_0^\pi \sin \left (\left (n + \frac{1}{2} \right ) t \right ) f(t) \, dt = 0,$$ or $$\lim_{n \to \infty} \left [\int_0^\pi \frac{\sin (n + \frac{1}{2}) t}{t} \, dt - \int_0^\pi \frac{\sin (n + \frac{1}{2}) t}{2 \sin (t/2)} \, dt \right ] = 0.\tag3$$
Now the second integral appearing in (3) can be found. From the Dirichlet kernel, namely $$\frac{1}{2} + \sum_{k = 1}^n \cos (kx) = \frac{\sin (n + \frac{1}{2}) x}{2 \sin (\frac{x}{2})},$$ we see that $$\int_0^\pi \frac{\sin (n + \frac{1}{2}) t}{2 \sin (\frac{t}{2})} \, dt = \int_0^\pi \left [\frac{1}{2} + \sum_{k = 1}^n \cos (kt) \right ] \, dt = \frac{\pi}{2}.$$ So (3) becomes $$\lim_{n \to \infty} \left [\int_0^\pi \frac{\sin (n + \frac{1}{2}) t}{t} \, dt - \frac{\pi}{2} \right ] = 0,$$ or $$\lim_{n \to \infty} \int_0^\pi \frac{\sin (n + \frac{1}{2}) t}{t} \, dt = \frac{\pi}{2},$$ as required to show.