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?

$\frac{\pi}{4}-\frac{1}{2}\le {\int}_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\le \frac{\pi}{2}-1$

$\frac{\pi}{4}-\frac{1}{2}\le {\int}_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\le \frac{\pi}{2}-1$

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

Beginner2022-10-09Added 7 answers

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, ${\int}_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\phantom{\rule{thinmathspace}{0ex}}dx\ge \frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\int}_{0}^{1}\mathrm{arcsin}x\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{\pi}{2}-1)=\frac{\pi}{4}-\frac{1}{2}.$

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, ${\int}_{0}^{1}\frac{\mathrm{arcsin}x}{1+{x}^{8}}\phantom{\rule{thinmathspace}{0ex}}dx\ge \frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{\int}_{0}^{1}\mathrm{arcsin}x\phantom{\rule{thinmathspace}{0ex}}dx=\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}(\frac{\pi}{2}-1)=\frac{\pi}{4}-\frac{1}{2}.$

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