The improper integrals of $\frac{\sin t}{t}$ and $\left|\frac{\sin t}{t}\right|$ Why is the integral of $\int_{0}^{\infty} \frac{\sin t}{t}dt =\frac{\pi}{2}$ and $\int_{0}^{\infty}\left|\frac{\sin t}{t}\right|dt = +\infty$? I did not studied improper integral, but for the first integral I have this hint:

what is the importance of saying that $(1/sin(\theta / 2)) - 2/\theta $ is continuous?
could anyone explain for me please?
 A: Write
$$a_n=\int_{(n-1)\pi}^{n\pi}\frac{\sin t}t\,dt.$$
Then for odd $n$ the integrand here is positive and for even $n$
it is negative. So $a_1$, $a_3,\ldots>0$ and $a_2$, $a_4,\ldots<0$.
By the alternating series test $\sum_n a_n$ converges, and it makes
sense to think of $\int_0^\infty\frac{\sin t}t\,dt$ as this
infinite sum.
But the convergence is conditional: $\sum_n|a_n|$ diverges. So
we can consider that $$\int_0^\infty\left|\frac{\sin t}t\right|\,dt
=\sum_{n=1}^\infty|a_n|=\infty.$$
A: $\int_{n\pi}^{(n+1)\pi}|\frac{sin t}{t}| dt> \frac 1{n+1}\int_{n\pi}^{(n+1)\pi}|\sin t| dt = \frac {2}{n}$
$\int_{n\pi}^{(n+1)\pi}|\frac {\sin t}{t}|dt > \sum_\limits{n=1}^{\infty} \frac {2}{n}$
Why does 
$\int_{0}^{\infty} \frac{\sin t}{t}dt$ converge?  The simple answer is that compares to an alternating series.
How to find the exact answer?  The method I know uses differentiation under the integral.
$f(a) = \int_{0}^{\infty} e^{-at}\frac{\sin t}{t}dt\\
f'(a) = \int_{0}^{\infty} -e^{-at} \sin t dt\\
f'(a) = \frac {e^{-at}(\cos t - a\sin t)}{1+a^2}|_0^{\infty}\\ 
f'(a) = -\frac {1}{1+a^2}\\
f(\infty) - f(0) =  -\int_0^{\infty} \frac {1}{1+a^2} da \\
f(\infty) - f(0) =  - \arctan \infty + \arctan 0 = -\frac {\pi}{2} \\
f(\infty) = \int_{0}^{\infty} 0\ dt = 0\\
f(0) = \int_{0}^{\infty} \frac{\sin t}{t}dt = \frac {\pi}{2}$
A: For tackling both problems one may employ integration by parts: $\sin(x)$ has mean zero but $\left|\sin x\right|$ has mean $\frac{2}{\pi}$. In particular:
$$ \int_{0}^{M}\frac{\sin x}{x}\,dx = \frac{1-\cos M}{M}+\int_{0}^{M}\frac{1-\cos x}{x^2}\,dx $$
gives that $\lim_{M\to +\infty}\int_{0}^{M}\frac{\sin x}{x}$ is finite, since $\frac{1-\cos x}{x^2}$ is an entire function and $0\leq \frac{1-\cos x}{x^2}\leq\frac{2}{x^2}$ for any $x\geq 1$. On the other hand, for any $k\in\mathbb{N}^+$
$$ \int_{0}^{k\pi}\frac{|\sin x|}{x}\,dx = \int_{0}^{\pi}\left|\sin x\right|\left(\frac{1}{x}+\frac{1}{x+\pi}+\ldots+\frac{1}{x+(k-1)\pi}\right)\,dx$$
is a positive integral that is lower-bounded by
$$ \int_{\pi/3}^{2\pi/3}\frac{\sqrt{3}}{2}\left(\frac{1}{\frac{2\pi}{3}}+\frac{1}{\frac{2\pi}{3}+\pi}+\ldots+\frac{1}{\frac{2\pi}{3}+(k-1)\pi}\right)\,dx $$
that is a quantity going to $+\infty$ like $C\log k$.
