Is it possible to evaluate $\int_0^1 \sin(\frac{1}{t})\,dt\,$? I was wandering if it possible to evaluate the value of the following improper integral:
$$
\int_0^1 \sin\left(\frac{1}{t}\right)\,dt
$$
It is convergent since  $\displaystyle\int_0^1 \left|\sin\left(\frac{1}{t}\right)\right|\,dt\leq \int_0^1 \;dt$, but I don't know if it is possible to calculate its value.
 A: Substituting $t = 1/x$ shows that the integral equals
$$
I=\int_1^\infty \frac{\sin x}{x^2} dx.
$$
Using integration by parts or otherwise, the indefinite integral can be shown to be
$$
\text{Ci}(x)-\frac{\sin (x)}{x},
$$
where
$$
\text{Ci}(x) = -\int_x^\infty \frac{\cos y}{y} dy
$$
is the Cosine Integral.
Plugging in the limits gives
$$
I = \sin 1 - \text{Ci}(1).
$$
Unfortunately this is not elementary.
A: Since, through a change of variable and integration by parts:
$$\int_{0}^{1}\sin\left(\frac{1}{t}\right)\,dt = \int_{1}^{+\infty}\frac{\sin t-\sin 1}{t^2}\,dt + \sin 1 = \sin 1+\int_{1}^{+\infty}\frac{\cos t}{t}\,dt$$
we just need to compute:
$$\int_{1}^{+\infty}\frac{\cos t}{t}\,dt = \sum_{k=0}^{+\infty}\int_{1+k\pi}^{1+(k+1)\pi}\frac{\cos t}{t}\,dt=\int_{1}^{\pi+1}\cos t\sum_{k=0}^{+\infty}\frac{(-1)^k}{t+k\pi}\,dt $$
that is:
$$\frac{1}{2\pi}\int_{1}^{\pi+1}\cos(t)\left(\psi\left(\frac{\pi+t}{2\pi}\right)-\psi\left(\frac{t}{2\pi}\right)\right)\,dt $$
and we are left with the integral of a smooth function over a compact interval.
