How to integrate to find each function and its limit (if it exists)? Find the expression for following three functions and then evaluate their limit.
$$f(x,y) = \int\limits_y^x-\cos({\pi t})\ln{t}\quad dt\qquad\qquad x\gt y$$
$\mathbf 1.$
$$\lim\limits_{x\to\infty} f(x,1)$$
$\mathbf 2.$
$$\lim\limits_{n\to\infty} f(n+1,n)\qquad n\in\mathbb{N}$$
$\mathbf 3.$
$$\lim\limits_{n\to\infty} f(n,n-1)\qquad n\in\mathbb{N}$$
 A: This integral cannot be evaluated using elementary functions, but using the Sine Integral, we get
$$
\begin{align}
\int_1^x\cos(\pi t)\log(t)\,\mathrm{d}t
&=\frac1\pi\sin(\pi x)\log(x)-\int_1^x\frac{\sin(\pi t)}{\pi t}\mathrm{d}t\\
&=\frac1\pi\left[\sin(\pi x)\log(x)-\operatorname{Si}(\pi x)+\operatorname{Si}(\pi)\right]
\end{align}
$$
However, we can show that this diverges without using anything complicated.
$$
\begin{align}
\int_{2n-\frac12}^{2n+\frac32}\cos(\pi t)\log(t)\,\mathrm{d}t
&=\int_{2n-\frac12}^{2n+\frac12}\cos(\pi t)\overbrace{\left[\log(t)-\log(t+1)\right]}^{\text{negative, increasing}}\,\mathrm{d}t\\
&\le\frac2\pi\log\left(\frac{2n+\frac12}{2n+\frac32}\right)\\
&\le-\frac4\pi\frac1{4n+3}
\end{align}
$$
Thus, the integral diverges by comparison with the Harmonic Series.

$$
\begin{align}
\int_n^{n+1}\cos(\pi t)\log(t)\,\mathrm{d}t
&=(-1)^n\int_0^1\cos(\pi t)\log(n+t)\,\mathrm{d}t\\
&=(-1)^n\int_0^1\cos(\pi t)\left[\log(n)+\log\left(1+\frac tn\right)\right]\,\mathrm{d}t\\
&=(-1)^n\int_0^1\cos(\pi t)\left[\log(n)+\left(\frac tn+O\left(\frac1{n^2}\right)\right)\right]\,\mathrm{d}t\\
&=(-1)^{n+1}\frac2{n\pi^2}+O\left(\frac1{n^2}\right)
\end{align}
$$

Thus, the sum of the integrals on the intervals $[n,n+1]$ is convergent; that is, it is the sum of two convergent series. However, the integral itself is not convergent.
A: For the first question, robjohn gave a nice answer.
Considering the second and third questions (which are the same) $$f(x,y) = \int\limits_y^x-\cos({\pi t})\log(t)\, dt=-\frac{-\text{Si}(\pi  x)+\text{Si}(\pi  y)+\log (x) \sin (\pi  x)-\log (y) \sin (\pi
    y)}{\pi }$$ where appears the sine integral. So $$f(n+1,n)=-\frac{\text{Si}(n \pi )-\text{Si}((n+1) \pi )-\log (n) \sin (\pi  n)+\log (n+1) \sin
   (\pi  (n+1))}{\pi }$$ but this simplifies since $n$ is an integer. So,$$f(n+1,n)=\frac{\text{Si}((n+1) \pi )-\text{Si}(n \pi )}{\pi }$$ Now, concerning the asymptotics of the sine integral for large arguments, I suggest you look here and you will easily conclude.
