Hope Hancock

2022-10-08

Prove the integral inequality
$\frac{\pi }{4}-\frac{1}{2}\le {\int }_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\le \frac{\pi }{2}-1$
$\frac{\pi }{4}-\frac{1}{2}\le {\int }_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\le \frac{\pi }{2}-1$
I can say that I have close to zero experience with integral inequalities. I've searched for couple of theorems most of them involving monotonous function on [0,1]. Arcsin is increasing in this case and the other one is decreasing. You could say we have a product of increasing and decreasing function on this interval however I didn't find any theorems that cover this case. Also this integral doesn't seem to have elementary function.
Another thing I noticed is that most of the theorems cover only 1 part of this inequality; either $\le or\ge$ so my guess is we need some type of combination or perhaps find a new integral that is equal to $\frac{\pi }{4}-\frac{1}{2}$ and another that is equal to $\frac{\pi }{2}-1$.
What's the correct approach when we need to prove that the value of certain integral is in a provided interval?

Ricky Lamb

Step 1
For the bound on the right, ${\int }_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\phantom{\rule{thinmathspace}{0ex}}dx\le {\int }_{0}^{1}\mathrm{arcsin}x\phantom{\rule{thinmathspace}{0ex}}dx=\frac{\pi }{2}-1.$.
Step 2
For the bound on the left, ${\int }_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\phantom{\rule{thinmathspace}{0ex}}dx\ge \frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\int }_{0}^{1}\mathrm{arcsin}x\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}\left(\frac{\pi }{2}-1\right)=\frac{\pi }{4}-\frac{1}{2}.$

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