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Background: A friend once brought up the integral $\displaystyle\int_0^\infty\frac{1}{1+x^4}\,dx$ and asked me if I can solve it. Now, I converted it into a beta function and solved it using Euler's reflection property (Evaluated to $\frac{\pi}{2\sqrt2}$). But my friend and I are in high school and he didn't know the use of beta function or most of the techniques that I had used there. He then challenged me to solve it using elementary high school metheds.

My Efforts: I thought that since I already knew the value it was going to evaluate to, I could brute-force my way to find this. I thought to use the sandwich theorem and take the bounds such that they evaluate to $\frac{\pi}{2\sqrt2}$.

For the upper bound, I thought that since I want $\pi$ in it I could use the derivative of $\arctan(x)$ since it is also similar to function we are trying to integrate.

$$\int_0^\infty \frac{1}{\sqrt2} \frac{1}{(x^2 + 1)}\,dx \geq\int_0^\infty \frac{1}{x^4+1} $$

Now, the problem here is that I can't exactly prove that this inequality is true.

I can definitely say that after a point $x\approx 1.3$ , $\displaystyle \frac{1}{\sqrt2}\frac{1}{1+x^2} > \frac{1}{1+x^4}$. But then from this I can't really comment on the inequality of the above integrals.

Also I can't find a suitable lower bound for this.

My Question:

Can we prove this inequality mentioned above, also what would be a suitable lower bound for this? Otherwise are there any better bounds I can use here (all knowing what the integral is going to converge to already) ? Or should I accept defeat as there is "no way" to solve this using elementary high school methods?

Any input appreciated..

Saket Gurjar
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    You can factorise $x^4+1=(x^2-\sqrt2 x+1)(x^2+\sqrt2 x+1)$ and use partial fractions. – Angina Seng Jun 25 '20 at 14:28
  • @MaximilianJanisch it does answer my question, but the reason I was looking to do it via the method mentioned in the question, is because I wanted to generalize it for $\int_0^\infty \frac{1}{1+x^n},dx$ – Saket Gurjar Jun 25 '20 at 14:32
  • I think the case of $\frac1{1+t^n}$ has also been covered on the site. – Jam Jun 25 '20 at 14:32
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    @SaketGurjar Have a look at this one for $\frac1{1+x^n}$: https://math.stackexchange.com/questions/781186/ I think doing partial fractions on that is too tough :) – Maximilian Janisch Jun 25 '20 at 14:34
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    You may write $$\frac{1}{x^4+1} = \frac{1}{2}\left(\frac{1+x^{-2}}{(x-x^{-1})^2+2} - \frac{1-x^{-2}}{(x+x^{-1})^2-2}\right). $$ The right-hand side can be easily integrated by substituting $u=x-x^{-1}$ and $v=x+x^{-1}$. – Sangchul Lee Jun 25 '20 at 14:42
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    I wrote complete and extremely detailed "high school level" calculations for closed form expressions of $\int \frac{1}{1 + x^n} dx$ for $n = 4, 5, 6$ in the early 2000s (used to be on the internet, but no longer seem to be). I can send .pdf files of these if you wish, but I'll need your email address. See my profile for my email address. Regarding the situation for an arbitrary positive integer $n,$ see my answer to Computing the primitive $\int\frac{1}{1+x^n} dx$. – Dave L. Renfro Jun 25 '20 at 17:16
  • If you know complex analysis, then for even values $n = 2m$ this is easy using contour integrals. It is half the integral from $-\infty$ to $+\infty$, which is in turn equal to $2\pi i$ times the sum of the residues in the upper half-plane. The residue corresponding to $\alpha = e^{(k+1/2)\pi i/m}$ is $1/2m\alpha^{2m-1} = \alpha/2m$, so the original integral is $\frac{\pi i}{2m} e^{\pi i/2m} \sum_{k = 0}^{m-1} e^{k \pi i/m} = \pi /[2m \sin (\pi/2m)]$. For a more elementary proof, it looks like the best option is partial fractions - obviously it will help to know complex numbers. – Anonymous Jun 25 '20 at 23:55
  • It should be obvious to you that your method of trying to squeeze the function has no chance of working. If $f$ and $g$ are continuous on $I = [0, +\infty)$, they have the same integral on $I$, and satisfy $f \geq g$, then we must in fact have $f = g$. So in your case, the fact that $1/[\sqrt{2}(x^2 + 1)]$ has the "right" integral guarantees that your inequality can't be right. – Anonymous Jun 25 '20 at 23:59

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