Is $\int_1^2 \frac{\cos t}{1-\sqrt t} \, dt$ integrable? Is $\displaystyle \int_1^2 \frac{\cos t}{1-\sqrt t} \, dt < \infty $ ? How to prove it ?
Thankfully
 A: HINT:
$$\frac{\cos(t)}{1-\sqrt{t}}=\frac{(1+\sqrt{t})\cos(t)}{1-t}$$
SPOILER ALERT:  SCROLL OVER THE HIGHLIGHTED AREA TO REVEAL THE SOLUTION

 Note that for for any $\epsilon >0$ and $x\in (1+\epsilon,3/2]$ we have $$\frac{(1+\sqrt{t})\cos(t)}{1-t}\le \frac{\left(1+\sqrt{\frac32}\right)\cos(3/2)}{1-t}$$and $$\int_{1+\epsilon}^{3/2}\frac{1}{1-t}\,dt=\log(2\epsilon)\to -\infty$$as $\epsilon \to 0^+$.  So, the integral of interest diverges.

A: You should check what happens near $1^+$:
$$\frac{\cos t}{1-\sqrt t}=\frac{\cos t}{1-\sqrt {1+(t-1)}}\sim
\frac{\cos 1}{1-(1+\frac{1}{2}(t-1))}=-\frac{2\cos 1}{t-1}$$
which is NOT integrable. Moreover $\displaystyle \int_1^2 \frac{\cos t}{1-\sqrt t} \, dt = -\infty$ because $-2\cos 1<0$
A: Assuming that the integral is convergent, the integral
$$ \int_{1}^{\sqrt{2}}\frac{2z}{1-z}\,\cos(z^2)\,dt $$
is convergent, too, due to the substitution $t=z^2$. However, the last integrand function is a meromorphic function with a simple pole at $z=1$ with residue $-2\cos(1)$, hence the last integral is not convergent and neither it is the original one.
