1. ## Improper Integral

Is there any method to evaluate $\displaystyle \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$?

If there is , please just say yes and hide the solution

and i think this integral should converge to a value :
$\displaystyle \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

$\displaystyle = \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} + \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

but $\displaystyle \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} = \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

and $\displaystyle \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }} < \int_{1}^{\infty} \frac{dx}{x^2}$ which converges to 1

Thus , the integral should be smaller than 2

2. Originally Posted by simplependulum
Is there any method to evaluate $\displaystyle \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$?

If there is , please just say yes and hide the solution

and i think this integral should converge to a value :
$\displaystyle \int_{0}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

$\displaystyle = \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} + \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

but $\displaystyle \int_{0}^{1} \frac{dx}{ \sqrt{ 1 + x^4 }} = \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }}$

and $\displaystyle \int_{1}^{\infty} \frac{dx}{ \sqrt{ 1 + x^4 }} < \int_{1}^{\infty} \frac{dx}{x^2}$ which converges to 1

Thus , the integral should be smaller than 2
yeah there is

$\displaystyle \frac{4\Gamma\left(\tfrac 54\right)^2}{\sqrt{\pi}}$