Gregory Jones

2022-01-05

I have in trouble for evaluating following integral
${\int }_{0}^{\mathrm{\infty }}\left(\sqrt{1+{x}^{4}}-{x}^{2}\right)dx=\frac{{\mathrm{\Gamma }}^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi }}$

star233

Let $x=\sqrt{\mathrm{tan}\theta }$. Note that, ${x}^{4}={\mathrm{tan}}^{2}\theta$ and $dx=\frac{{\mathrm{sec}}^{2}\theta d\theta }{2\sqrt{\mathrm{tan}\theta }}$ Like this,
$I={\int }_{0}^{\mathrm{\infty }}\left(\sqrt{1+{x}^{4}}-{x}^{2}\right)dx={\int }_{0}^{\pi /2}\frac{\left(\mathrm{sec}\theta +\mathrm{tan}\theta \right){\mathrm{sec}}^{2}\theta d\theta }{2\sqrt{\mathrm{tan}\theta }}=\frac{1}{2}{\int }_{0}^{\pi /2}\frac{{\mathrm{cos}}^{1/2}\theta \left(1-\mathrm{sin}\theta \right)d\theta }{{\mathrm{sin}}^{1/2}\theta {\mathrm{cos}}^{3}\theta }$
$=\frac{1}{2}{\int }_{0}^{\pi /2}{\mathrm{cos}}^{-5/2}\theta {\mathrm{sin}}^{-1/2}\theta d\theta -\frac{1}{2}{\int }_{0}^{\pi /2}{\mathrm{cos}}^{-5/2}\theta {\mathrm{sin}}^{1/2}d\theta ={I}_{1}+{I}_{2}$
Note that, ${I}_{1}=\frac{1}{2}B\left(-3/4,1/4\right)$, because

But, $B\left(m,n\right)=\frac{1}{2}\frac{\mathrm{\Gamma }\left(m\right)\mathrm{\Gamma }\left(n\right)}{\mathrm{\Gamma }\left(m+n\right)}$. Thus,
${I}_{1}=\frac{1}{4}\frac{\mathrm{\Gamma }\left(-3/4\right)\mathrm{\Gamma }\left(1/4\right)}{\mathrm{\Gamma }\left(-1/2\right)}=\frac{1}{4}\cdot \frac{\left(-4\right)}{3}\frac{\mathrm{\Gamma }\left(1/4\right)\cdot \mathrm{\Gamma }\left(1/4\right)}{\left(-2\right)\sqrt{\pi }}=\frac{\mathrm{\Gamma }\left(1/4{\right)}^{2}}{6\sqrt{\pi }}$
If $x\to 0,\mathrm{\Gamma }\left(x\right)\to \mathrm{\infty }$, so that ${I}_{2}=0$. Furthermore, if define $\mathrm{\Gamma }\left(x\right)=\mathrm{\Gamma }\left(x+1\right)/x$

user_27qwe

Another approach :
Set
${x}^{4}=\frac{1}{t}-1⇒x=\left(\frac{1-t}{t}{\right)}^{\frac{1}{4}}⇒dx=-\frac{1}{4}{t}^{-\frac{5}{4}}\left(1-t{\right)}^{-\frac{3}{4}}dt$
Then the integral turms out to be
${\int }_{0}^{\mathrm{\infty }}\left(\sqrt{1+{x}^{4}}-{x}^{2}\right)dx=\frac{1}{4}{\int }_{0}^{1}\left({t}^{-\frac{7}{4}}\left(1-t{\right)}^{-\frac{3}{4}}-{t}^{-\frac{7}{4}}\left(1-t{\right)}^{-\frac{1}{4}}\right)dt\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[B\left(-\frac{3}{4},\frac{1}{4}\right)-B\left(-\frac{3}{4},\frac{3}{4}\right)\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\left[B\left(-\frac{3}{4},\frac{1}{4}\right)-0\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{4}\cdot \frac{\mathrm{\Gamma }\left(-\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)}{\mathrm{\Gamma }\left(-\frac{1}{2}\right)}\phantom{\rule{0ex}{0ex}}=\frac{{\mathrm{\Gamma }}^{2}\left(\frac{1}{4}\right)}{6\sqrt{\pi }}$

Vasquez

Let's apply the substitution $\sqrt{{x}^{4}+1}-{x}^{2}=\sqrt{t}$: