<s>[INST] Find a lower bound for a Riemann integral. Let $a,b$ be  strictly positive constants and let $p\in(1/4,3/4)$. Is there $c>0$ such that
$$\int_0^1(1-t^a)^{1/2+p}[2t-(2+b)t^{b+1}]dt>c \quad ?$$
*The constant $c>0$ may depend on $a,b$ or $p$.
I tried several values for $(a,b,p)$, and always observed a positive integral. So, I'm conjecturing that the integral is strictly positive.
Intuitively, $2t-(2+b)t^{b+1}$ is a concave function on $[0,1]$ which is weighted by a factor $(1-t^a)^{1/2+p}$. This factor is less than one, and decreases as $t\to 1$, and so it pushes the right tail of the function (which is negative) closer to zero. So it makes sense to think about a positive integral.
I wrote the integral in terms of Beta functions but didn't succeed.
 [/INST] Firstly, note that
\begin{align}
\int_0^1\lvert 1-t^a\rvert^{1/2+p}\lvert 2t-(2+b)t^{b+1}\rvert dt\leq\int_0^12t-(2+b)t^{b+1}=2<\infty.
\end{align}
Suppose that $1/4<p<1/2$. Now, using the Fubini's Theorem and the Binomial Theorem for the fractional exponent $q:=1/2+p$ ,
\begin{align}
&\int_0^1(1-t^a)^q[2t-(2+b)t^{b+1}]dt=\int_0^1\sum_{k=0}^\infty \binom{q}{k}(-t^a)^k[2t-(2+b)t^{b+1}]dt\\
=&\sum_{k=0}^\infty \binom{q}{k}(-1)^k \int_0^1\ t^{ak}[2t-(2+b)t^{b+1}]dt=\sum_{k=0}^\infty \binom{q}{k}(-1)^k\bigg[\frac{2}{ak+2}-\frac{2+b}{ak+2+b}\bigg]\\
=&\sum_{k=1}^\infty \binom{q}{k}(-1)^{k+1}\bigg[\frac{bak}{(ak+2)(ak+2+b)}\bigg]\\
=&\sum_{k\text{ odd}} \binom{q}{k}\bigg[\frac{bak}{(ak+2)(ak+2+b)}\bigg]-\binom{q}{k+1}\bigg[\frac{ba(k+1)}{(a(k+1)+2)(a(k+1)+2+b)}\bigg] \\
=&\sum_{k\text{ odd}} \binom{q}{k}ab\bigg[\frac{k}{(ak+2)(ak+2+b)}+\frac{(k-q)}{(ak+a+2)(ak+a+2+b)}\bigg]  \ *obs. (k-q\geq k-1)\\
\geq & \sum_{k\text{ odd}} \binom{q}{k}ab\bigg[\frac{ak(2ak^2+ak+6k+2bk+a)+2a+4k+2bk}{(ak+2)(ak+2+b)(ak+a+2)(ak+a+2+b)}\bigg] \quad *obs. \binom{q}{k}\geq 0,\text{if } k \text{ odd}\\
>&\binom{q}{1}ab\bigg[\frac{a(2a+a+6+2b+a)+2a+4+2b}{(a+2)(a+2+b)(a+a+2)(a+a+2+b)}\bigg]>0.
\end{align}
where $\binom{q}{k}=q(q-1)\dotsc(q-k+1)/k!$.
When $1/2\leq p\leq 3/4$, similar argument can be applied rewriting the sum by means of even indices.
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