Problem with integral Is there a number $1<y<\infty$ such that
$$
\int_{1}^\infty \frac{1}{t} \frac{\sin t}{t} dt =\int_y ^\infty \frac{\sin t}{t} dt  \textrm{ ? }
$$
 A: This is a numerical answer.
The RHS is the negative of the (lowercase) sine integral, $\int_y ^\infty \frac{\sin t}{t} dt= -\mathbf {si}(y)$ for which we have $-\mathbf {si}(y) = \frac\pi2 -\mathbf {Si}(y) $ , $\mathbf {Si}(y)$ being the (uppercase) sine integral. Now, in the interval $(1,\infty)$ we have $\min \mathbf {Si}(y) \approx 0.95$ and $\max \mathbf {Si}(y) \approx 1.85$. So the corresponding range of $-\mathbf {si}(y)$ is $[\frac\pi2 -1.85 , \frac\pi2 -0.95] \approx [-0.28, 0.62]$. Then the question can be reformulated as:
Does 
$$\int_{1}^\infty \frac{1}{t} \frac{\sin t}{t} dt \in [-0.28, 0.62] \textrm{ ? }$$
Calculation gives 
$$\int_{1}^\infty \frac{1}{t} \frac{\sin t}{t} dt \approx 0.50$$
so the answer to the question is, yes, and $y \in (1.15,1.16)$.
A: Yes. Let $$\int_{1}^\infty \frac{1}{t} \frac{\sin t}{t} dt = C.$$ Now $$f(y) = \int_y ^\infty \frac{\sin t}{t} dt$$ is a continuous function of $y$, $f(1) \geq C$, $f(y) \to 0$ as $y\to\infty$, and $C > 0$. 
There should be such a $y \in (0,\pi)$, I believe, since $$\int_{\pi}^\infty \frac{1}{t} \frac{\sin t}{t} dt < 0.$$
Edit: Thomas Andrews has pointed out correctly that it's not clear $f(1)$ is $\geq C$.
