Estimation of some Integrals, which exceeded Wolfram Alpha's free computation time.

Question

I want to have an upper bound for $$\int_1^\frac{1}{p}(1-sp)^\frac{\gamma}{p} \frac{1-\log(s)}{s}ds-\int_1^\infty e^{-\gamma s}\frac{1-\log(s)}{s}ds.$$ However, I could not compute either one of these integrals with wolfram Alpha. The bound can be as high as possible but it should be constant. Note that $p \in (0,1)$ and $\gamma \in (0,\infty)$. I do not need an actual antiderivative

Answer 1

Observe that $1-x-e^{-x}$ has derivative $-(1-e^{-x})$, so it has a global maximum at $x=0$ and is decreasing for $x>0$. Then

$$0<1-sp < e^{-sp}$$

and hence

$$(1-sp)^{1/p} < e^{-s} \implies (1-sp)^{\gamma/p} < e^{-\gamma \,s}.$$

Now, rewrite your expression as

$$ \underbrace{\int_1^\frac{1}{p}\, \underbrace{\left((1-sp)^\frac{\gamma}{p} - e^{-\gamma\,s} \right)}_{<0}\frac{1-\log(s)}{s}\,ds}_{(\triangle)} \quad \underbrace{- \int_{1/p}^\infty\, e^{-\gamma s}\,\frac{1-\log(s)}{s}\,ds}_{(*)}.$$

We further expand $(*)$

$$ \underbrace{\int_{1/p}^\infty\, e^{-\gamma s}\,\frac{\log(s)}{s}\,ds}_{>0} \quad\underbrace{- \int_{1/p}^\infty\, e^{-\gamma s}\,\frac{1}{s}\,ds}_{<0} $$

and hence need only consider the first term above. We write

$$\int_{1/p}^\infty\, e^{-\gamma s}\,\frac{\log(s)}{s}\,ds < \int_{1/p}^\infty\, e^{-\gamma s}\,\log(s)\,ds < \int_{1/p}^\infty\, e^{-\gamma s}\,s\,ds = \frac{(p+\gamma)\,e^{-\gamma/p}}{p\,\gamma^2}. $$

It remains only to bound $(\triangle)$. Write:

\begin{align} (\triangle) &= \underbrace{\int_1^\frac{1}{p}\, \left((1-sp)^\frac{\gamma}{p} - e^{-\gamma\,s} \right)\frac{1}s\,ds}_{<0} \quad - \int_1^\frac{1}{p}\, \left((1-sp)^\frac{\gamma}{p} - e^{-\gamma\,s} \right)\frac{\log(s)}{s}\,ds \\&< \int_1^\frac{1}{p}\, \left( e^{-\gamma\,s} - (1-sp)^\frac{\gamma}{p} \right)\frac{\log(s)}{s}\,ds \\&< \int_1^\frac{1}{p}\, e^{-\gamma\,s}\,\frac{\log(s)}{s}\,ds \\&< \int_1^\frac{1}{p}\, e^{-\gamma\,s}\,s\,ds < \int_0^\frac{1}{p}\, e^{-\gamma\,s}\,s\,ds = \frac{p-(p+\gamma)\,e^{-\gamma/p}}{p\,\gamma^2}. \end{align}

Combining the two bounds, we find that regardless of the value of $p$ your initial expression is less than $\gamma^{-2}$.