For which values of k does the function is bounded? The exact question says "for what values of the positive real number $k$ are the functions $f(x)$ bounded for $x\approx 0^+$ ?"


*

*a) $$\int_x^1 \frac{1}{t^k}dt$$

*b) $$\int_x^1 \frac{e^t}{t^k}dt$$


I´ve been struggling with this question for a couple of days. For the first one, I got the integral, which is $\frac{1-x^{1-k}}{1-k}$, and then tried to evaluate it by $x\approx 0^+$ but I got stuck.
$$\frac{1-0_+^{1-k}}{1-k}$$
For the second one, I have no idea how to proceed as the integral is too much complicated. Any advise how to proceed will be useful.
 A: Hint. Concerning the first integral you have found that
$$
0<k<1 \tag1
$$ one may then observe that
$$
e^t=1+O(t),\quad t\ge0,
$$ giving
$$
\frac{e^t}{t^k}=\frac{1}{t^k}+O\left(t^{1-k}\right) \tag2
$$ thus the second integral is convergent for $0<k<1$ too.
A: For $(a)$, if $k=1$, we have that $f(x)=\ln(x)-\ln(1)=\ln(x)$. As $x\to0^+$, $f(x)\to-\infty$, so it is not bounded in this case.
If $k=0$, then $f(x)=x-1$. As $x\to0^+$, $f(x)\to -1$ and it is bounded in this case.
If $k$ is neither $0$ nor $1$, then $f(x)=\int_1^xt^{-k}\,dt=\left[\frac{1}{1-k}t^{1-k}\right]_1^x=\frac{1}{1-k}\cdot(x^{1-k}-1)$. Now $\lim_{x\to0^+}f(x)$ depends entirely on $\lim_{x\to0^+}x^{1-k}$. We see that $f$ is unbounded if $1-k<0$, that is $k>1$, and bounded if $1-k>0$.

All in all, for $(a)$, $f$ is unbounded near $0^+$ if and only if $k\geq 1$.


For $(b)$, notice that on $[0,1]$ we have that $e^t\geq 1$, so $\frac{e^t}{t^k}\geq \frac1{t^k}$. It follows that $f$ is unbounded when $k\geq 1$, from the previous result.
On the other hand, on $[0,1]$, $e^t\leq e<3$, so $\frac{e^t}{t^k}<3\frac1{t^k}$. It follows that $f$ is bounded when $k<1$, again from the previou result.
