6

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

involves the Gamma function. I think that the generalised form $$ I_n=\int_0^{\infty}\left(\sqrt{1+x^{2 n}}-x^n\right) d x $$ could be expressed in terms of the Gamma function too for any natural number $n$ greater than $1$.

Now putting $x\mapsto \frac{1}{x}$ helps us to use integration by parts.

$$ \begin{aligned} I _n & =\int_0^{\infty}\left(\sqrt{1+\frac{1}{x^{2 n}}}-\frac{1}{x^n}\right) \frac{d x}{x^2} \\ & =\int_0^{\infty} \frac{\sqrt{x^{2 n}+1}-1}{x^{n+2}} d x \\ & =-\frac{1}{n+1} \int_0^{\infty}\left(\sqrt{x^{2 n}+1}-1\right) d\left(\frac{1}{x^{n+1}}\right) \\ & =\frac 1 {n+1} \int_0^{\infty} \frac{1}{x^{n+1}} \cdot \frac{2 n x^{2 n-1}}{2 \sqrt{x^{2 n}+1}} \\ & =\frac{n}{n+1} \underbrace{\int_0^{\infty} \frac{x^{n-2}}{\sqrt{x^{2 n}+1}} d x}_{J_n} \end{aligned} $$

Putting $ y=\frac 1{x^{2n}+1} $ converts the integral into a Beta function.

$$ \begin{aligned} J_n & =\frac{1}{2 n} \int_0^1 y^{\frac{1}{2}}\left(\frac{1}{y}-1\right)^{\frac{n-2}{2 n}}\left(\frac{1}{y}-1\right)^{\frac{1}{2 n}-1} \frac{d y}{y^2} \\ & =\frac{1}{2 n} B\left(\frac{1}{2 n},-\frac{1}{2 n}+\frac{1}{2}\right) \end{aligned} $$

Plugging $J_n$ back yields $$ \begin{aligned} I_n & =\frac{1}{2(n+1)} B\left(\frac{1}{2 n},-\frac{1}{2 n}+\frac{1}{2}\right) \\ & =\frac{1}{2(n+1)} \frac{\Gamma\left(\frac{1}{2 n}\right) \Gamma\left(-\frac{1}{2 n}+\frac{1}{2}\right)}{\Gamma\left(\frac{1}{2}\right)} \\ \end{aligned} $$ Now we can generalise it as $$\boxed{\int_0^{\infty}\left(\sqrt{1+x^{2 n}}-x^n\right) d x =\frac{1}{2(n+1) \sqrt{\pi}} \Gamma\left(\frac{1}{2 n}\right) \Gamma\left (\frac{1}{2} -\frac{1}{2 n}\right) }$$

For examples:

$$ \begin{aligned} & I_3=\frac{1}{8 \sqrt{\pi}} \Gamma\left(\frac{1}{6}\right) \Gamma\left(\frac{1}{3}\right) \\ & I_8=\frac{1}{18 \sqrt{\pi}} \Gamma\left(\frac{1}{16}\right) \Gamma\left(\frac{7}{16}\right) \end{aligned} $$

checked by WA.

Latest Edit

As I look back after submitting the post, I found that the proof of my generalisation doesn’t need that $n$ is natural number greater than $1$ and hence the generalisation is true for any real number $a>1$ i.e.

$$\boxed{\int_0^{\infty}\left(\sqrt{1+x^{2 a}}-x^a\right) d x =\frac{1}{2(a+1) \sqrt{\pi}} \Gamma\left(\frac{1}{2 a}\right) \Gamma\left (\frac{1}{2} -\frac{1}{2 a}\right) }$$

For example:

$$ \int_0^{\infty}\left(\sqrt{1+x^{2 e}}-x^e\right) d x=\frac{1}{2(e+1) \sqrt{\pi}} \Gamma\left(\frac{1}{2 e}\right) \Gamma\left(\frac{1}{2}-\frac{1}{2 e}\right) $$ checked by WA.

My Question:

Are there any other generalisation of $\int_0^{\infty}\left(\sqrt{1+x^{4}}-x^2\right) d x $?

Lai
  • 31,615

3 Answers3

10

Consider $$\displaystyle \int_0^{\infty}x^{s-1}\left(\sqrt{1+x^{2 n}}-x^n\right)^t d x$$

\begin{eqnarray*} \int_0^{\infty}x^{s-1}\left(\sqrt{1+x^{2 n}}-x^n\right)^t d x &=& \frac{1}{n}\int_0^{\infty}w^{\frac{s}{n}-1}\left(\sqrt{1+w^2}-w\right)^t d w \quad \left(w \mapsto x^n\right)\\ &=& \frac{1}{2^{\frac{s}{n}}n}\int_0^{1}(1-v^2)^{\frac{s}{n}-1}v^{t-\frac{s}{n}-1}(v^2+1) d v \quad \left(v \mapsto \sqrt{1+w^2}-w\right)\\ &=& \frac{1}{2^{\frac{s}{n}}n}\int_0^{1}(1-v^2)^{\frac{s}{n}-1}v^{t-\frac{s}{n}+1} d v + \frac{1}{2^{\frac{s}{n}}n}\int_0^{1}(1-v^2)^{\frac{s}{n}-1}v^{t-\frac{s}{n}-1} d v\\ &=& \frac{1}{2^{\frac{s}{n}+1}n}\int_0^{1}(1-y)^{\frac{s}{n}-1}y^{\frac{t}{2}-\frac{s}{2n}} d y + \frac{1}{2^{\frac{s}{n}+1}n}\int_0^{1}(1-y)^{\frac{s}{n}-1}y^{\frac{t}{2}-\frac{s}{2n}-1} d y \quad \left(y \mapsto v^2\right)\\ &=& \frac{t}{2^{\frac{s}{n} }(tn+s)}\frac{\Gamma\left(\frac{s}{n}\right)\Gamma\left(\frac{t}{2}-\frac{s}{2n}\right)}{\Gamma\left(\frac{t}{2}+\frac{s}{2n}\right)}\\ \end{eqnarray*}

$$ \boxed{\int_0^{\infty}x^{s-1}\left(\sqrt{1+x^{2 n}}-x^n\right)^t d x = \frac{t}{2^{\frac{s}{n} }(tn+s)}\frac{\Gamma\left(\frac{s}{n}\right)\Gamma\left(\frac{t}{2}-\frac{s}{2n}\right)}{\Gamma\left(\frac{t}{2}+\frac{s}{2n}\right)} \quad s<tn \quad s,t,n\in \mathbb{R}_{> 0}}$$

Bertrand87
  • 3,977
3

As @Gonçalo suggested $$\int\left(\sqrt[n]{1+x^4}-\sqrt[n]{x^4}\right)dx=x \,\, _2F_1\left(\frac{1}{4},-\frac{1}{n};\frac{5}{4} ;-x^4\right)-\frac{x^{\frac{4}{n}+1}}{\frac{4}{ n}+1}$$

If $n>2$ $$\int_0^{\infty}\left(\sqrt[n]{1+x^4}-\sqrt[n]{x^4}\right)dx=\Gamma \left(\frac{5}{4}\right) \frac{\Gamma \left(-\frac{1}{n}-\frac{1}{4}\right)}{\Gamma\left(-\frac{1}{n}\right)}$$ works for any $n$

$$\int_0^{\infty}\left(\sqrt[n]{1+x^5}-\sqrt[n]{x^5}\right)dx=\Gamma \left(\frac{6}{5}\right) \frac{\Gamma \left(-\frac{1}{n}-\frac{1}{5}\right)}{\Gamma\left(-\frac{1}{n}\right)}$$

$$\int_0^{\infty}\left(\sqrt[n]{1+x^6}-\sqrt[n]{x^6}\right)dx=\Gamma \left(\frac{7}{6}\right) \frac{\Gamma \left(-\frac{1}{n}-\frac{1}{6}\right)}{\Gamma\left(-\frac{1}{n}\right)}$$ The pattern is obvious

Claude Leibovici
  • 289,558
1

We can readily generalize in the same manner as you did but with the additional parameter $m$: \begin{align} &\int_0^\infty (1+x^{2n})^{1/m}-x^{2n/m}\ \mathrm dx=\int_0^\infty \frac{(1+x^{2n})^{1/m}-1}{x^{2n/m+2}}\mathrm dx,\qquad [x\rightarrow 1/x]\\\\ =&\frac{2n}{2n+m}\int_0^\infty \frac{(1+x^{2n})^{1/m-1}}{x^{2n(1/m-1)+2}}\mathrm dx,\qquad \text{by parts}\\\\ =&\frac{1}{2n+m}\int_0^1x^{1/2n-1}(1-x)^{-1/2n-1/m}\mathrm dx,\qquad \left[\frac{1}{1+x^{2n}}\rightarrow x\right]\\\\ =&\frac{1}{2n+m}B\left(\frac{1}{2n},1-\frac{1}{2n}-\frac 1m\right)=\frac{1}{2n+m}\frac{\Gamma\left(\frac{1}{2n}\right)\Gamma\left(1-\frac{1}{2n}-\frac 1m\right)}{\Gamma\left(1-\frac 1m\right)}. \end{align} Numerical approximations indicate that for $m,n\in\mathbb R_{>1}$ the integral converges.

Eli Bartlett
  • 2,084