Show that $$ \int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} $$ converges.

I recognized that that since the integrand is even then $$ \int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} = 2\int_{0}^{\infty}\frac{dx}{\sqrt{x^4+1}} $$

Then, I started with the following statement and manipulated it to match the integrand. $$ \begin{align*} x^4 +1 &\gt x^4 \\ \sqrt{x^4+1} &\gt x^2 \\ \frac{1}{\sqrt{x^4 +1}} &\lt\frac{1}{x^2} \end{align*} $$

It is a well-known result that $$ \int_1^{\infty}\frac{dx}{x^2} $$ converges. So would it be correct to say that, by comparison, that

$$ \int_{-\infty}^{\infty}\frac{dx}{\sqrt{x^4+1}} $$ converges as well?

  • 2
    $\begingroup$ Not quite. Notice how the lower limit of your integrand has changed from 0 to 1. $\endgroup$ – Sinister Cutlass Jan 29 '16 at 0:42
  • $\begingroup$ Since $\frac1{\sqrt{x^4+1}}\le\frac1{x^2}$ on $[1,\infty)$ and $\frac1{\sqrt{x^4+1}}\le1$ on $[0,1)$, you can compare on the proper intervals, then use symmetry. $\endgroup$ – robjohn Nov 8 '18 at 15:41

That this integral converges should be clear from the $1/x^2$ behavior as $x \to \pm \infty$. We can see the convergence in action as we evaluate the integral. Here, we demonstrate how to transform the integral into something more familiar by using complex analytic techniques.

Consider the following contour integral:

$$\oint_C \frac{dz}{\sqrt{1+z^4}} $$

where $C$ is the following contour:

enter image description here

The idea is to avoid the branch points. Thus, I made $C$ a semicircle in the upper half place of radius $R$, with circular detours about the branch points at $z=e^{i \pi/4}$ and $z=e^{i 3 \pi/4}$. The contour integral is then equal to

$$\int_{-R}^R \frac{dx}{\sqrt{1+x^4}} + i R \int_0^{\pi} d\theta \, \frac{e^{i \theta}}{\sqrt{1+R^4 e^{i 4 \theta}}} \\ + e^{i \pi/4} \int_R^{1+\epsilon} \frac{dt}{\sqrt[+]{1-t^4}} + i \epsilon \int_{\pi/4}^{9 \pi/4} d\phi \, \frac{e^{i \phi}}{\sqrt{1+\epsilon^4 e^{i 4 \phi}}} \\ + e^{i \pi/4} \int_{1+\epsilon}^R \frac{dt}{\sqrt[-]{1-t^4}} + e^{i 3 \pi/4} \int_R^{1+\epsilon} \frac{dt}{\sqrt[+]{1-t^4}} \\ + i \epsilon \int_{3 \pi/4}^{11 \pi/4} d\phi \, \frac{e^{i \phi}}{\sqrt{1+\epsilon^4 e^{i 4 \phi}}} + e^{i 3 \pi/4} \int_{1+\epsilon}^R \frac{dt}{\sqrt[-]{1-t^4}}$$

Note that $\sqrt[+]{1-t^4} = i \sqrt{t^4-1}$ represents the positive branch of the square root, while $\sqrt[-]{1-t^4} = -i \sqrt{t^4-1}$ represents the negative branch of the square root.

To recover the real integral we want, we take the limits as $R \to \infty$ and $\epsilon \to 0$. In this limit, the second, fourth, and seventh integrals vanish. (This is straightforward to verify in this case and I will not work that out here.) We are left with, as the contour integral in these limits:

$$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{1+x^4}} +i 2 \left (e^{i \pi/4} + e^{i 3 \pi/4} \right ) \int_1^{\infty} \frac{dt}{\sqrt{t^4-1}} $$

Now, this is where I will strike at a misconception: we do not use the residue theorem here. The reason is that there are no poles enclosed within the contour $C$. Rather, we use Cauchy's theorem (of which the residue theorem is a special case), which states that the contour integral is zero because the integrand is analytic within the contour $C$. Now, using the fact that $e^{i \pi/4} + e^{i 3 \pi/4} = i \sqrt{2}$ and using the symmetry of the integrand, we get the following:

$$\int_{0}^{\infty} \frac{dx}{\sqrt{1+x^4}} = \sqrt{2} \int_1^{\infty} \frac{dt}{\sqrt{t^4-1}} $$

So we have used Cauchy's theorem to transform one integral into another. Before I make remarks about what this means, let's evaluate the integral on the RHS, which is a lot easier then the original integral. First, note that

$$\int_1^{\infty} \frac{dt}{\sqrt{t^4-1}} = \int_0^1 \frac{dt}{\sqrt{1-t^4}} $$

Now sub $t=u^{1/4}$, then $dt = \frac14 u^{-3/4} du$, and we now have

$$\int_{0}^{\infty} \frac{dx}{\sqrt{1+x^4}} = \frac{\sqrt{2}}{4} \int_0^1 du \, u^{-3/4} (1-u)^{-1/2} $$

The latter integral is a beta function. Thus, we have an analytical result for the integral:

$$\int_{0}^{\infty} \frac{dx}{\sqrt{1+x^4}} = \frac{\sqrt{2}}{4} \frac{\displaystyle \Gamma \left (\frac14\right ) \Gamma \left (\frac12\right )}{\displaystyle \Gamma \left (\frac34\right )} $$

Use the facts that $\Gamma \left (\frac12\right ) = \sqrt{\pi}$, and

$$\Gamma \left (\frac34\right ) \Gamma \left (\frac14\right ) = \frac{\pi}{\sin{(\pi/4)}} = \sqrt{2} \pi$$

Putting this altogether, we finally have

$$\int_{-\infty}^{\infty} \frac{dx}{\sqrt{1+x^4}} = \frac1{2 \sqrt{\pi}} \Gamma \left (\frac14\right )^2 $$


In the regime $|x| \ge 1$, you can compare $(x^4+1)^{-1/2}$ against $x^{-2}$, but in the neighborhood of $0$, the latter is unbounded so you are best off noting that $$\int_{x=-\infty}^\infty (x^4+1)^{-1/2} \, dx = 2 \int_{x=0}^\infty (x^4+1)^{-1/2} \, dx < 2\left( \int_{x=0}^1 (x^4+1)^{-1/2} \, dx + \int_{x=1}^\infty x^{-2} \, dx \right).$$ The first term is obviously finite since $0 < (x^4 + 1)^{-1/2} \le 1$ on $[0,1]$; and the second is also finite as you stated.


basically, yes. It's symmetric even, so we only need to care the positive part. It's smooth so the only issue is convergence at infinity and you have shown that.


$$ \begin{align} \int_{-\infty}^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}} &=2\int_0^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag1\\ &=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_1^\infty\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag2\\ &=2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}+2\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag3\\ &=4\int_0^1\frac{\mathrm{d}x}{\sqrt{x^4+1}}\tag4 \end{align} $$ Explanation:
$(1)$: symmetry
$(2)$: split the integral into two parts
$(3)$: substitute $x\mapsto\frac1x$
$(4)$: distribute

Because the integrand is between $\frac1{\sqrt2}$ and $1$, the integral converges and is between $2\sqrt2$ and $4$.

  • $\begingroup$ Hi @robjohn! I have a big rep lately! Finally, a good decision here since the great questions make a big difference. $\endgroup$ – user 1591719 Dec 15 '19 at 14:48

Your approach works well, but keep in mind that the convergence of$\newcommand{\dd}{{\rm d}}$

$$\int_{1}^{\infty} {\frac{1}{x^2}} \: \dd x$$

only implies that

$$\int_{1}^{\infty} {\frac{1}{\sqrt{1+x^4}}} \: \dd x$$

converges by comparison. However you can extend this further because clearly

$$\int_{0}^{1} {\frac{1}{\sqrt{1+x^4}}} \: \dd x$$

converges. Therefore

$$\int_{0}^{\infty} {\frac{1}{\sqrt{1+x^4}}} \: \dd x$$

converges, and by extension

$$2\int_{0}^{\infty} {\frac{1}{\sqrt{1+x^4}}} \: \dd x$$



$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x \over \root{1 + x^{4}}}} = 2\int_{0}^{\infty}{\dd x \over \root{1 + x^{4}}}}$. Let $\ds{t \equiv {1 \over 1 + x^{4}}}$ such that $\ds{x = \pars{{1 \over t} - 1}^{1/4}}$.


\begin{align} \ds{\color{#f00}{\int_{-\infty}^{\infty}{\dd x \over \root{1 + x^{4}}}}} & = 2\int_{1}^{0}t^{1/2}\,{1 \over 4}\,\pars{{1 \over t} - 1}^{-3/4} \pars{-\,{1 \over t^{2}}}\,\dd t \\[3mm] & = \half\int_{0}^{1}t^{-3/4}\pars{1 - t}^{-3/4}\,\dd t = \half\,{\Gamma\pars{1/4}\Gamma\pars{1/4} \over \Gamma\pars{1/4 + 1/4}} \\[3mm] & = \color{#f00}{{1 \over 2\root{\pi}}\,\Gamma^{2}\pars{{1 \over 4}}} \approx 3.7081 \end{align}

since $\Gamma\pars{1/2} = \root{\pi}$.

  • $\begingroup$ 1. The question is not to compute the integral but to show that it converges. 2. The "Beta" argument is already in Ron's answer posted 5 months ago. $\endgroup$ – Did May 19 '16 at 8:26

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