Prove $\int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx=\frac{\Gamma^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi}}$

Solution 1:

Let's apply the substitution $\sqrt{x^4+1}-x^2=\sqrt{t}$: $$x^2=\frac{1-t}{2\sqrt{t}},\quad dx=-\frac{\sqrt2}{8}t^{-5/4}(t+1)(1-t)^{-1/2}dt,$$ $$\int_0^\infty\left(\sqrt{x^4+1}-x^2\right)\,dx=\frac{\sqrt2}{8}\left(\int_0^1 t^{1/4}(1-t)^{-1/2}\,dt+\int_0^1 t^{-3/4}(1-t)^{-1/2}\,dt\right)=$$ $$=\frac{\sqrt2}{8}\left(\mathrm{B}\left(\frac54,\frac12\right)+\mathrm{B}\left(\frac14,\frac12\right)\right)=\frac{\Gamma^2(\frac14)}{6\sqrt{\pi}}.$$

Solution 2:

Another approach :

Set $$ x^4=\frac1t-1\quad\color{blue}{\Rightarrow}\quad x=\left(\frac{1-t}t\right)^{\large \frac14}\quad\color{blue}{\Rightarrow}\quad dx=-\frac14t^{-\large\frac54}\left(1-t\right)^{-\large\frac34}\ dt, $$ then the integral turns out to be \begin{align} \int_0^{\infty} \left(\sqrt{1+x^{4}}-x^{2}\right)\ dx&=\frac14\int_0^1 \left(t^{-\large\frac74}\left(1-t\right)^{-\large\frac34}-t^{-\large\frac74}\left(1-t\right)^{-\large\frac14}\right)\ dt\\ &=\frac14\left[\text{B}\left(-\frac34,\frac14\right)-\text{B}\left(-\frac34,\frac34\right)\right]\\ &\stackrel{\color{red}{[1]}}=\frac14\left[\text{B}\left(-\frac34,\frac14\right)-\color{red}0\right]\\ &\stackrel{\color{red}{[2]}}=\frac14\cdot\frac{\Gamma\left(-\frac34\right)\Gamma\left(\frac14\right)}{\Gamma\left(-\frac12\right)}\\ &\stackrel{\color{red}{[3]}}=\color{blue}{\frac{\Gamma^2\left(\frac14\right)}{6\sqrt\pi}}. \end{align}

Notes :

$\color{red}{[1]}\;\;\;$Use property of Beta function and Euler's reflection formula for the gamma function $$\text{B}(x,y)=\text{B}(x+1,y)+\text{B}(x,y+1)$$ $\quad\;\;\;$and $$\Gamma(z)\Gamma(1-z)=\frac{\pi}{\sin\pi z}$$ $\color{red}{[2]}\;\;\;\displaystyle\text{B}(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$

$\color{red}{[3]}\;\;\;\displaystyle\Gamma(n)=\frac{\Gamma(n+1)}n$ and $\displaystyle\Gamma\left(\frac12\right)=\sqrt\pi$

Solution 3:

Let $x = \sqrt{\tan \theta}$. Note that, $x^4 = \tan^2 \theta$ and $dx = \dfrac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}$. Like this, $$ I = \int_{0}^{\infty}( \sqrt{1 + x^4} - x^2)dx = \int_{0}^{\pi/2} \dfrac{(\sec \theta + \tan \theta)\sec^2\theta d\theta}{2\sqrt{\tan \theta}} = \dfrac{1}{2}\int_{0}^{\pi/2} \dfrac{\cos^{1/2}\theta (1 - \sin \theta)d\theta}{\sin^{1/2}\theta \cos^3 \theta} $$ $$ = \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{-1/2}\theta d\theta - \dfrac{1}{2}\int_{0}^{\pi/2}\cos^{-5/2}\theta\sin^{1/2}\theta d\theta = I_1 + I_2 $$ Note that, $I_1 = \dfrac{1}{2}B(-3/4,1/4)$, because $$ (2m-1=-5/2 \ \Rightarrow \ m = -3/4 \ \text{and} \ 2n-1 = -1/2 \ \Rightarrow \ n=1/4) $$ But, $B(m,n) = \dfrac{1}{2}\dfrac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}$. Thus, $$ I_1 = \dfrac{1}{4}\dfrac{\Gamma(-3/4)\Gamma(1/4)}{\Gamma(-1/2)} = \dfrac{1}{4}\cdot \dfrac{(-4)}{3}\dfrac{\Gamma(1/4)\cdot \Gamma(1/4)}{(-2)\sqrt{\pi}} = \dfrac{\Gamma(1/4)^2}{6\sqrt{\pi}} $$ If $x \to 0$, $\Gamma(x) \to \infty$, so that $I_2 = 0$. Furthermore, if $x < 0, x \neq -1,-2,\ldots$, define $\Gamma(x) = \Gamma(x + 1)/x$.