Find a lower bound for a Riemann integral.

Question

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.

Answer 1

$$\int_0^1 (1-t^a)^{1/2+p}[2t-(2+b)t^{b+1}]~dt$$ $$=\int_0^1 (1-t^a)^{1/2+p}~d(t^2-t^{b+2})$$ $$=\left.(1-t^a)^{1/2+p}(t^2-t^{b+2})\right|_0^1-\int_0^1(t^2-t^{b+2})(\frac 1 2+p)(1-t^a)^{p-1/2}(-at^{a-1})~dt$$ $$=0+\int_0^1(t^2-t^{b+2})(\frac 1 2+p)(1-t^a)^{p-1/2}(at^{a-1})~dt,\qquad (*)$$ which is positive. The latter integral is bounded by $$\max_{0\leq t\leq 1}(t^2-t^{b+2})\int_0^1(\frac 1 2+p)(1-t^a)^{p-1/2}\cdot at^{a-1}~dt=\max_{0\leq t\leq 1}(t^2-t^{b+2}),$$ where one uses the substitution $u=t^a$ for computing the last integral. It follows that when $b\rightarrow 0$, the integral approaches $0.$

By the above argument, there does not exist a uniform positive lower bound. However, one may still obtain a lower bound dependent on $a,b,$ and $p$:

Let $f(t)=t^2-t^{b+2}=t^2(1-t^b)\geq 0$. Then $f$ has a unique critical point at $t=\left(\frac 2{b+2}\right)^{1/b}$ in $(0,1)$ and $f$ is increasing on $[1/2,(2/(b+2))^{1/b}]$ (Note that $2^{b+1}>b+2$ for $b>0$ as can be checked by the fact that $g(x)=2^{x+1}-x-2$ satisfies $g(0)$ and $g'(x)>0$ for all $x\geq 0$). It follows from $(*)$ that $$\int_0^1 (1-t^a)^{1/2+p}[2t-(2+b)t^{b+1}]~dt$$ $$=\int_0^1 f(t)(\frac 1 2+p)(1-t^a)^{p-1/2}(at^{a-1})~dt$$ $$>\int_{1/2}^{(2/(b+2))^{1/b}} f(t)(\frac 1 2+p)(1-t^a)^{p-1/2}(at^{a-1})~dt$$ $$>f(1/2)\int_{1/2}^{(2/(b+2))^{1/b}} (\frac 1 2+p)(1-t^a)^{p-1/2}(at^{a-1})~dt$$ $$=\left[\left(\frac 1 2\right)^2-\left(\frac 1 2\right)^{b+2}\right]\cdot \left[\left.-(1-t^a)^{p+1/2}\right|_{1/2}^{(2/(b+2))^{1/b}}\right]$$ $$=\left[\left(\frac 1 2\right)^2-\left(\frac 1 2\right)^{b+2}\right]\cdot\left[\left(1-\left(\frac 1 2\right)^a\right)^{p+1/2}-\left(1-\left(\frac 2{b+2}\right)^{a/b}\right)^{p+1/2}\right],$$ hence a lower bound can be taken to be $$c(a,b,p):=\left[\left(\frac 1 2\right)^2-\left(\frac 1 2\right)^{b+2}\right]\cdot\left[\left(1-\left(\frac 1 2\right)^a\right)^{p+1/2}-\left(1-\left(\frac 2{b+2}\right)^{a/b}\right)^{p+1/2}\right]>0.$$

Answer 2

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.