I have in trouble for evaluating following integral \int_0^\infty(\sqrt{1+x^4}-x^2)dx=\frac{\Gamma^2(\frac{1}{4})}{6\sqrt{\pi}}

Gregory Jones 2022-01-05
I have in trouble for evaluating following integral
\(\displaystyle{\int_{{0}}^{\infty}}{\left(\sqrt{{{1}+{x}^{{4}}}}-{x}^{{2}}\right)}{\left.{d}{x}\right.}={\frac{{\Gamma^{{2}}{\left({\frac{{{1}}}{{{4}}}}\right)}}}{{{6}\sqrt{{\pi}}}}}\)

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star233
Answered 2022-01-11 Author has 0 answers

Let \(x=\sqrt{\tan\theta}\). Note that, \(x^4=\tan^2\theta\) and \(dx=\frac{\sec^2\theta d\theta}{2\sqrt{\tan\theta}}\) Like this,
\(I=\int_0^\infty(\sqrt{1+x^4}-x^2)dx=\int_0^{\pi/2}\frac{(\sec\theta+\tan\theta)\sec^2\theta d\theta}{2\sqrt{\tan\theta}}=\frac{1}{2}\int_0^{\pi/2}\frac{\cos^{1/2}\theta(1-\sin\theta)d\theta}{\sin^{1/2}\theta\cos^3\theta}\)
\(=\frac{1}{2}\int_0^{\pi/2}\cos^{-5/2}\theta\sin^{-1/2}\theta d\theta-\frac{1}{2}\int_0^{\pi/2}\cos^{-5/2}\theta\sin^{1/2}d\theta=I_1+I_2\)
Note that, \(I_1=\frac{1}{2}B(-3/4,1/4)\), because
\((2m-1=-5/2\Rightarrow m=-3/4\ and\ 2n-1=-1/2\Rightarrow n=1/4)\)
But, \(B(m,n)=\frac{1}{2}\frac{\Gamma(m)\Gamma(n)}{\Gamma(m+n)}\). Thus,
\(I_1=\frac{1}{4}\frac{\Gamma(-3/4)\Gamma(1/4)}{\Gamma(-1/2)}=\frac{1}{4}\cdot\frac{(-4)}{3}\frac{\Gamma(1/4)\cdot\Gamma(1/4)}{(-2)\sqrt{\pi}}=\frac{\Gamma(1/4)^2}{6\sqrt{\pi}}\)
If \(x\to0,\Gamma(x)\to\infty\), so that \(I_2=0\). Furthermore, if \(x<0,\ x\ne-1,-2,..., \)define \(\Gamma(x)=\Gamma(x+1)/x\)

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user_27qwe
Answered 2022-01-11 Author has 9558 answers

Another approach :
Set
\(x^4=\frac{1}{t}-1\Rightarrow x=(\frac{1-t}{t})^{\frac{1}{4}}\Rightarrow dx=-\frac{1}{4}t^{-\frac{5}{4}}(1-t)^{-\frac{3}{4}}dt\)
Then the integral turms out to be
\(\int_0^\infty(\sqrt{1+x^4}-x^2)dx=\frac{1}{4}\int_0^1(t^{-\frac{7}{4}}(1-t)^{-\frac{3}{4}}-t^{-\frac{7}{4}}(1-t)^{-\frac{1}{4}})dt \\=\frac{1}{4}[B(-\frac{3}{4},\frac{1}{4})-B(-\frac{3}{4},\frac{3}{4})] \\=\frac{1}{4}[B(-\frac{3}{4},\frac{1}{4})-0] \\=\frac{1}{4}\cdot\frac{\Gamma(-\frac{3}{4})\Gamma(\frac{1}{4})}{\Gamma(-\frac{1}{2})} \\=\frac{\Gamma^2(\frac{1}{4})}{6\sqrt{\pi}}\)

0
Vasquez
Answered 2022-01-11 Author has 9055 answers

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

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