TITLE: Estimate $\int_0^{\infty} 1/\sqrt{1+x^4} \mathrm{d}x$
QUESTION [6 upvotes]: I need an analytical estimation of the following integral:
$$\int\limits_0^\infty \frac{{\mathrm{d} x}}{\sqrt{1 + x^4}}$$
It has a root in the denomenator -- so I can't make use of complex residues technique.
Edit:
Since CAS can do it symbolically -- there's certainly a solution. However I did this analytically a couple of years ago and obtained 
$$    \frac{\pi}{2^{3/4}}$$
estimation. I remember it was easy and fast estimation. I'm trying to recover it.

REPLY [4 votes]: For $x \in [0,\infty]$, let $I(x)$ be the integral $\;\displaystyle\int_0^x \frac{dt}{\sqrt{1+t^4}}$. 
The integral we want to calculate is simply $I(\infty)$. 
Quoting some results from this answer
of a related question, we know for $x \in [0,\infty)$,
$$I(x) = F(\sqrt{1+x^2-\sqrt{1+x^4}}; \frac12 )$$
where
$$F(y\,; m) = \int_0^y \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}$$
is the Jacobi's form of 
incomplete elliptic integral of the first kind.
When $x \to \infty$, $\sqrt{1 + x^2 - \sqrt{1+x^4}} \to 1$. This leads to
$$I(\infty) = K\left(\frac12\right)\quad\text{ where }\quad 
K(m) = \int_0^1 \frac{ds}{\sqrt{(1-s^2)(1-ms^2)}}$$
is the complete elliptic integral of the first kind.
$K(m)$ can be computed efficiently using its relation with the arithmetic geometric mean of $1$ and $\sqrt{1-m}$.
$$K(m) = \frac{\pi}{2\text{AGM}(1,\sqrt{1-m})}$$
Start with $m = \frac12$, the first iteration of computing the AGM gives us
$$ \frac{1}{\sqrt[4]{2}} = \text{GM}(1,\frac{1}{\sqrt{2}} ) \le \text{AGM}(1,\frac{1}{\sqrt{2}} ) \le \text{AM}(1, \frac{1}{\sqrt{2}} ) = \frac12 (1 + \frac{1}{\sqrt{2}})$$
and hence 
$$ 1.840302369 \sim \frac{\sqrt{2}\pi}{\sqrt{2}+1} \le I(\infty) = K\left(\frac12\right) \le \frac{\pi}{2^{3/4}} \sim 1.868002168 $$
To get more accurate estimate of the integral, one can carry out more iterations in the AGM
computation. For example, the second iteration gives us
$$ 1.854048814 \sim
\frac{2^{3/2}\pi }{2^{5/4}+2^{1/2}+1}
\le K\left(\frac12\right) \le 
\frac{\pi }{\sqrt{2^{1/4} + 2^{3/4}}}
\sim 1.8541005407$$
which is accurate to about $4^{th}$ decimal places.