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

I have in trouble for evaluating following integral

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

It seems really easy, but I don't know how to handle it at all. (The results are well known, here I tried to evaluate it but I failed)

I tried to use the relation

$$\sqrt{1+x^{4}}-x^{2}=\frac{1}{\sqrt{1+x^{4}}+x^{2}}$$

but I couldn't find the desired results.

• Have you looked at the substitution $u\mapsto x^2$? That seems like it would transform this into a relatively classical and manageable elliptic integral... Jul 15, 2014 at 6:37
• Hint: the value you're looking at is $${1\over 6}B({1\over 4},{1\over 4}).$$ Jul 15, 2014 at 6:49
• This is extremely similar to my own question. Jul 15, 2014 at 6:49

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

• Out of the three posted "solutions," this is the only one that is correct. +1 Aug 21, 2016 at 4:10

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 $\displaystyle\Gamma\left(\frac12\right)=\sqrt\pi$

• Isn't Euler's beta function defined as corresponding integral only for positive arguments? Jul 15, 2014 at 18:00
• This development is not rigorous since one may not split the integral into the difference of two divergent integrals. With some modifications, the development can be made rigorous. Aug 21, 2016 at 4:09

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 $\Gamma(x) = \Gamma(x + 1)/x$.

• Both $I_1$ and $I_2$ diverge as $\theta\rightarrow \pi/2$. Jul 15, 2014 at 7:43
• I did not pay attention to the divergence of integrals. I agree with your observation, but the strange thing is that I got the above result. The other post above also includes negative binomial. I do not know how to circumvent the problem of divergence. Thanks for the comment. Jul 15, 2014 at 8:00
• One way is to introduce an extra parameter $s$ so that $\cos^{-5/2}\theta$ become $\cos^{-5/2+s}\theta$. This will give two analogs of $I_1$ and $I_2$ convergent for $\Re s>3/2$ and expressible in terms of gamma functions. Then one may continue analytically in $s$ (to $s=0$) using the recurrence relations for $\Gamma(z)$ and the fact that the result has no singularities in a larger domain $\Re s > -\frac12$. Jul 15, 2014 at 8:08
• Interesting technique. I would like to see it in detail. Jul 15, 2014 at 8:17
• Thank you, It seems beta function is powerful tools for evaluating such integrals, i try to be familiar with this tool. Jul 15, 2014 at 9:12