I have found in my advanced analysis book an exercise which asks about finding certain estimate of the Incomplete Gamma function. Concretely, it reads as follows:
The Incomplete Gamma function $\Gamma(a,x)$ is defined as $\Gamma(a,x)=\int_x^\infty t^{a-1}e^{-t}$. If $1<a<2$ and $0\leq s \leq 1$, could you find a constant $C$ such that $\Gamma(a,-\ln(s))\leq Cs$?
I am a little lost and I don't know where to start with this exercise. Could someone help me?
The way that the question is written is making it more difficult. For short let $$F(a,s)=\Gamma(a,-\log(s))$$ I'll leave it to you to check (exercise!!) that $$(\partial_sF)(a,s)=(-\log(s))^{a-1}$$ So, in the vicinty of $s=0$, the function looks like a vertical line, that is to say $(\partial_sF)$ is unbounded in the neighborhood of zero. Hence there is no linear upper bound in this region. To nail the point home, let's say we have found some $c$ such that $$F(a,s)\leq cs\\ \text{when} ~~0\leq s\leq 1$$ Let's pick some small number $s_0>0$. (Of course, we'd prefer to do our analysis at precisely $s_0=0$, but the derivative breaks in that case.) We know that $F(a,s_0)\geq 0$. And, we know that $$(\partial_sF)(a,s_0)=(-\log(s_0))^{a-1}=|\log(s_0)|^{a-1}$$ And therefore picking $\epsilon$ to be sufficiently small, we can say $$F(a,s_0+\epsilon)\geq \epsilon|\log(s_0)|^{a-1}$$ If we compare this to our upper bound, which is $$c(s_0+\epsilon)$$ It is pretty clear no matter how large or small $\epsilon$ is we can always find an $s_0$ such that $$\epsilon|\log(s_0)|^{a-1}\geq c(s_0+\epsilon)$$ Play with this other graph to reinforce your intuition.
Maybe this idea is a little unclear, but hopefully playing with the graph I've provided here and in the comments will help.