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$$ \int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x $$

I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and $(x)^2-1/(x)^2=z $ but no helpful expression was derived. I also used property $\int_0^a f(a-x)=\int_0^a f(x) $ Please help me out

Felix Marin
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Sourav
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6 Answers6

23

We can do better than hypergeometric function and elliptic integral: $$\color{blue}{\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{{{\Gamma ^2}(\frac{1}{4})}}{{12\sqrt \pi }}}$$

Firstly, integration by part gives $$\int_0^1 {\sqrt {1 + {x^4}} dx} = \sqrt 2 - 2\int_0^1 {\frac{{{x^4}}}{{\sqrt {1 + {x^4}} }}dx} = \sqrt 2 - 2\int_0^1 {\left( {\sqrt {1 + {x^4}} - \frac{1}{{\sqrt {1 + {x^4}} }}} \right)dx} $$ Hence $$\int_0^1 {\sqrt {1 + {x^4}} dx} = \frac{{\sqrt 2 }}{3} + \frac{2}{3}\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} $$ Making $x=1/u$ in the last integral gives $$\int_0^1 {\frac{1}{{\sqrt {1 + {x^4}} }}dx} = \frac{1}{2}\int_0^\infty {\frac{1}{{\sqrt {1 + {x^4}} }}} dx = \frac{1}{{8\sqrt \pi }}{\Gamma ^2}(\frac{1}{4})$$ which can be evaluated by using some formula for beta function.

pisco
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  • +1 cool. it is nice to see one can replace the hypergeometric function by the more elementary gamma function. – achille hui Jun 06 '17 at 06:37
  • This is same as Robert Israel 's evaluation but in a different way. The value $K(1/\sqrt{2})$ is evaluated in terms of Beta/Gamma function as $\Gamma^{2}(1/4)/(4\sqrt{\pi})$. – Paramanand Singh Jun 06 '17 at 06:39
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    See the evaluation of $K(1/\sqrt{2})$ at https://math.stackexchange.com/a/1793756/72031 – Paramanand Singh Jun 06 '17 at 06:45
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    Btw in general if $y^{2}=p(x)$ where $p(x) $ is a polynomial of degree $3$ or $4$ and $R(x, y) $ is a rational function of two variables then the integral $\int R(x, y) , dx$ is non-elementary and can be evaluating terms of elliptic integrals. So looking at the integrand $\sqrt{1+x^{4}}$ it is reasonable to expect elliptic integrals. But your technique appears to be very simple because it avoids the difficult theory of elliptic integrals. +1 already provided for your nice answer. – Paramanand Singh Jun 06 '17 at 06:51
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Consider the ${}_{2}F_{1}$ hypergeometric integral form given by $${}_{2}F_{1}(a, b; c; x) = \frac{\Gamma(c)}{\Gamma(b) \, \Gamma(c-b)} \, \int_{0}^{1} t^{b-1} \, (1-t)^{c-b-1} \, (1-x \, t)^{-a} \, dt$$ leads to, with $a=-1/2$, $b=1/4$, $c=5/4$, $x=-1$, $${}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = \frac{1}{4} \, \int_{0}^{1} t^{-3/4} \, \sqrt{1+ t} \, dt.$$ Now let $t = x^{4}$ to obtain $$\int_{0}^{1} \sqrt{1 + x^4} \, dx = {}_{2}F_{1}\left(-\frac{1}{2}, \frac{1}{4}; \frac{5}{4}; -1\right) = 1.08943...$$

Leucippus
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5

We can do better than elliptic integrals or $\Gamma$ / Beta / hypergeometric functions: $$ \int_{0}^{1}\sqrt{1+t^4}\,dt = \frac{\sqrt{2}}{3}+\frac{\pi}{6\,\text{AGM}\left(1,\frac{1}{\sqrt{2}}\right)} \tag{1}$$ due to integration by parts and what is shown in this answer. $(1)$ summarizes a very efficient numerical technique (the $\text{AGM}$ iteration has a quadratic convergence) for computing arbitrarily accurate numerical approximations of the LHS. It also shows $$ 1.084\ldots=\frac{\sqrt{2}}{3}+\frac{\pi}{3}(2-\sqrt{2})\leq \int_{0}^{1}\sqrt{1+t^4}\,dt\leq \frac{\sqrt{2}}{3}+\frac{\pi}{6}2^{1/4}=1.094\ldots\tag{2}$$ Improving the bound $\leq\sqrt{\frac{6}{5}}$ given by Jensen's inequality.
Your integral is strictly related to the lemniscate constant.

Jack D'Aurizio
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Integration by parts gives $$ \begin{aligned} I & =\int_0^1 \sqrt{1+x^4} d x \\ & =\int_0^1 \frac{1}{\sqrt{1+x^4}} d x+\int_0^1 \frac{x^4}{\sqrt{1+x^4}} d x \\ & =\int_0^1 \frac{1}{\sqrt{1+x^4}} d x+\frac{1}{2} \int_0^1 x d\left(\sqrt{1+x^4}\right) \\ & =\int_0^1 \frac{1}{\sqrt{1+x^4}} d x+\frac{1}{2}\left[x \sqrt{1+x^4}\right]_0^1-\frac{1}{2} \int_0^1 \sqrt{1+x^4} d x \\ \therefore I & =\frac{2}{3} \int_0^1 \frac{1}{\sqrt{1+x^4}} d x+\frac{\sqrt{2}}{3} \end{aligned} $$ For the first integral, letting $x^2=\tan \theta$ transforms it into $$ \begin{aligned} \int_0^1 \frac{1}{\sqrt{1+x^4}} d x & =\frac{1}{2} \int_0^{\frac{\pi}{4}} \frac{1}{\sqrt{\cos \theta \sin \theta}} d \theta \\ & =\frac{\sqrt{2}}{4} \int_0^{\frac{\pi}{2}} \sin ^{2\left(\frac{1}{4} \right) -1} \theta \cos ^{2\left(\frac{1}{2}\right)-1} d \theta\\&=\frac{\sqrt{2}}{8} B\left(\frac{1}{4}, \frac{1}{2}\right)\\&= \frac{1}{8 \sqrt{\pi}} \Gamma^2\left(\frac{1}{4}\right) \quad \left(\textrm{ By } \Gamma\left(\frac{1}{4}\right) \Gamma\left(\frac{3}{4}\right)=\pi \csc \left(\frac{\pi}{4}\right) \right)\end{aligned} $$ Hence $$\boxed{I=\frac{1}{12 \sqrt{\pi}} \Gamma^2\left(\frac{1}{4}\right) +\frac{\sqrt{2}}{3}} $$

Lai
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Another non-elementary answer, from Maple, is $$ \int_0^1 \sqrt{1+x^4}\; dx = \frac{\sqrt {2}+{\it EllipticK} \left( 1/\sqrt {2} \right)}{3} $$

Robert Israel
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For an approximation, you could use a Padé approximant for the integrand. The simplest one would be $$\frac{3 x^4+4}{x^4+4}=3-\frac{x+2}{x^2+2 x+2}+\frac{x-2}{x^2-2 x+2}$$ $$\int \frac{3 x^4+4}{x^4+4}\,dx=3x+\frac{1}{2} \log \left(\frac{x^2-2 x+2}{x^2+2 x+2}\right)+\tan ^{-1}(1-x)-\tan ^{-1}(1+x)$$ So, using the given bounds, an approximation is $$\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,dx \approx 3-\frac{\log (5)}{2}-\tan ^{-1}(2)\approx 1.08813$$

Claude Leibovici
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