Estimation of a defined integral

Question

I need to show that $$\left|\int\limits_{0}^{\frac{1}{2}}\frac{(2t-1)^3}{(\sqrt{1+t})^7}dt\right|<\frac{16}{125}$$

Evaluating it would be my last hope, but it wouldn't be easy wither. Is there a trick for that kind of problems?

Answer 1

The denominator is greater than 1. So the value is less than $\int_0^{0.5}(1-2t)^3dt=\frac{1}{8}=\frac{16}{128}<\frac{16}{125}$.

Answer 2

This is not an answer but it is too long for a comment.

Since your last hope is to evaluate the integral, you will be more than happy (I am sure !) to hear that $$8\left|\int\limits_{0}^{\frac{1}{2}}\frac{(2t-1)^3}{(\sqrt{1+t^7}}dt\right|=4 \, _2F_1\left(\frac{1}{7},\frac{1}{2};\frac{8}{7};-\frac{1}{128}\right)-6 \, _2F_1\left(\frac{2}{7},\frac{1}{2};\frac{9}{7};-\frac{1}{128}\right)+$$ $$4 \, _2F_1\left(\frac{3}{7},\frac{1}{2};\frac{10}{7};-\frac{1}{128}\right)-\, _2F_1\left(\frac{1}{2},\frac{4}{7};\frac{11}{7};-\frac{1}{128}\right)$$ where appear a bunch of Gaussian hypergeometric functions (see here).

The value of the rhs is $\approx 0.9999881703$