Test for convergence 
Possible Duplicate:
Does $ \int_0^{\infty}\frac{\sin x}{x}dx $ have an improper Riemann integral or a Lebesgue integral? 

I am stuck with the following integral:
\begin{equation}
\int_\mathbb{R} \frac{\sin t}{t}
\end{equation}
I would like to find out whether this integral is convergent, but I totally forgot how to find the right convergence test here, at the origin I should have no problem since the integrand has a removable singularity. Towards infinity I was thinking about the harmonic series as a comparison, but obviously it is not going to bound the integrand from below, which would show divergence . Integration by parts should give me a logarithm multiplied by a trigonometric function, would that help? I then have to figure out $$\lim_{t \to \infty} \cos t \log t $$ .. how do I do this?
For the absolute integrand I reckon it is easier, here I can bound below using the harmonic series. But for the integral above I am not sure, thks for any hints!
 A: Consider the "improper Riemman integral", 
\begin{align*}
\int_0^\infty {\sin(x)\over x} \,dx &= \lim_{n\to\infty} \int_0^{\pi n}{\sin(x)\over x} \,dx \\
&= \lim_{n \to \infty}\sum_{k=0}^{n -1} \int_{k\pi}^{(k+1)\pi}{\sin(x)\over x}\,dx \\
&=  \sum_{k = 0}^\infty (-1)^k
\int_0^\pi{\sin(x)\over x + k\pi} \,dx
\end{align*}
The integrals are decreasing in $k$, so the series converges at $n\to\infty$ by the alternating series test.  However, this integral is NOT absolutely integrable.  
A: A good idea to try whenever we have an integral with an oscillating factor and a dampening factor is to strengthen the dampening by integrating by parts. In this problem, behavior near $0$ is not the problem, so here it yields $$ \int^{\infty}_1 \frac{\sin x}{x} dx = \frac{- \cos x}{x} \biggr|^{\infty}_1 + \int^{\infty}_1 \frac{\cos x}{x^2} dx $$
which is certainly finite since the integral on the right hand side is absolutely convergent. 
A: This answer has been moved (and slightly modified) to an answer to a duplicate question.
The integral is not absolutely convergent. Because $\int_{k\pi}^{(k+1)\pi}|\sin(t)|\;\mathrm{d}t=2$, we have
$$
\frac{2}{(k+1)\pi}\le\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin(t)}{t}\right|\;\mathrm{d}t\le\frac{2}{k\pi}\tag{1}
$$
Since the harmonic series diverges, so does the integral of the absolute value.
However, the improper integral
$$
\lim_{N\to\infty}\int_{-N}^N\frac{\sin(t)}{t}\mathrm{d}t\tag{2}
$$
does exist. To see this, note that
$$
\int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\mathrm{d}t=0\tag{3}
$$
With $a=\frac12\left(\frac{1}{2k\pi}+\frac{1}{2(k+1)\pi}\right)$, recalling $\int_{2k\pi}^{2(k+1)\pi}|\sin(t)|\;\mathrm{d}t=4$, and using $(3)$, we get
$$
\begin{align}
\left|\int_{2k\pi}^{2(k+1)\pi}\frac{\sin(t)}{t}\mathrm{d}t\right|
&=\left|\int_{2k\pi}^{2(k+1)\pi}\sin(t)\;\left(\frac1t-a\right)\;\mathrm{d}t\right|\\
&\le4\cdot\frac12\left(\frac{1}{2k\pi}-\frac{1}{2(k+1)\pi}\right)\\
&=\frac{1}{k(k+1)\pi}\tag{4}
\end{align}
$$
and
$$
\sum_{k=1}^\infty\frac{1}{k(k+1)}=1\tag{5}
$$
Therefore, $(1)$, $(4)$, and $(5)$ guarantee that
$$
\int_{2\pi}^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{6}
$$
converges to a value no greater than $\dfrac1\pi$.
Since $\left|\dfrac{\sin(t)}{t}\right|\le1$,
$$
\int_0^{2\pi}\frac{\sin(t)}{t}\mathrm{d}t\tag{7}
$$
has a value no greater than $2\pi$.
Since $\dfrac{\sin(t)}{t}$ is even, $(6)$ and $(7)$ guarantee that
$$
\int_{-\infty}^\infty\frac{\sin(t)}{t}\mathrm{d}t\tag{8}
$$
converges to a value no greater than $4\pi+\dfrac2\pi$.

Another general test is Dirichlet's test (Theorem 17.5). In this case,
$$
\left|\int_0^N\sin(t)\;\mathrm{d}t\right|\le2
$$
and $\dfrac1t$ is monotonically decreasing to $0$ on $(0,\infty)$. Thus, by Dirichlet,
$$
\int_0^\infty\frac{\sin(t)}{t}\mathrm{d}t
$$
converges.

By symmetry,
$$
\int_{-\infty}^0\frac{\sin(t)}{t}\mathrm{d}t
$$
converges also.
