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Let $n\in\mathbb N$ ($n>0$), and define the $n$th integral in the sequence $I$ to be

$$I_n = \int_{-\infty}^{\infty}\frac{1}{1+x^{2n}}dx.$$

Evaluating such integrals, especially for small $n$, is essentially a straightforward exercise in complex analysis (integrate on a semicircle in the UHP using the residue theorem, send the radius to $+\infty$ and show that the integral on the "curved part" goes to zero). However, rather surprisingly the values of these integrals get progressively more non-trivial as $n$ increases:

\begin{align} \\I_1 &= \pi \approx 3.14 \\I_2 &= \frac{\pi}{\sqrt 2}\approx 2.22 \\I_3 &= \frac{2\pi}{3}\approx 2.09 \\I_4 &= \frac{\pi}{4}\csc\frac{\pi}{8}\approx 2.05 \\I_5 &= \frac{4\pi}{5(\sqrt 5 - 1)}\approx 2.033 \\I_6 &= \frac 1 6\Gamma\left(\frac 1 {12}\right)\Gamma\left(\frac{11}{12}\right)\approx 2.023 \\ &\dots \end{align}

Is anything currently known about:

$a)$ General closed forms for $I_n$,

$b)$ The stabilization of the numerical value of $I_n$ as $n\to\infty$? Noting that these integrals are bounded below by zero and that $I_n$ is a decreasing sequence, the limit $$\lim_{n\to\infty}I_n$$ must be finite. Can we say more (either numerically or analytically)?

Venus
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theage
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4 Answers4

4

You may recall the celebrated $\Gamma$ function defined by $$ \Gamma(\alpha)=\int_0^\infty u^{\alpha-1} e^{-u}\:{\rm{d}}u, \quad \alpha>0 $$ Observe that $$ \int_{-\infty}^\infty \frac{1}{x^{2n}+1} \:{\rm{d}}x=2\int_0^\infty \frac{1}{x^{2n}+1} \:{\rm{d}}x $$ then you may write $$ \begin{align} \int_0^\infty \frac{1}{x^{2n}+1} \:{\rm{d}}x&=\int_0^\infty\int_0^\infty e^{-(x^{2n}+1)t} \:{\rm{d}}t\:{\rm{d}}x \\&=\int_0^\infty e^{-t}\int_0^\infty e^{-x^{2n} t} \:{\rm{d}}t\:{\rm{d}}x \\&=\int_0^\infty e^{-t}\left(\int_0^\infty e^{-x^{2n} t}\:{\rm{d}}x\right)\:{\rm{d}}t \\&=\frac1{2n}\int_0^\infty t^{-\frac1n}e^{-t}\left(\int_0^\infty u^{\frac1{2n}-1} e^{-u}{\rm{d}}u\right)\:{\rm{d}}t \\&=\frac1{2n} \Gamma\left(1-\frac1{2n}\right)\Gamma\left(\frac1{2n}\right) \\&=\frac{\pi}{{2n}\sin \frac{\pi}{{2n}}} \end{align} $$ and $$ I_n=\frac{\pi}{{n}\sin \frac{\pi}{{2n}}} $$ which is easy to handle, you thus have $$\lim_{n\to\infty}I_n=2.$$

Olivier Oloa
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3

Since the integrand is an even function, we can rewrite $$I_n=2\int_0^\infty\frac{\mathrm dx}{1+x^{2n}}$$ and we have a well-known result for the latter integral, namely $$\int_0^\infty\frac{\mathrm dx}{1+x^{2n}}=\frac{\pi}{2n}\csc\left(\!\frac{\pi}{2n}\!\right)$$ Hence $$I_n=\frac{\pi}{n}\csc\left(\!\frac{\pi}{2n}\!\right)$$ and as $n\to\infty$, we get $$\lim_{n\to\infty}I_n=2$$

Community
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Venus
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2

Noting that $f_n\colon x\in\mathbb{R}\mapsto \frac{1}{1+x^{2n}}$ converges pointwise towards $f = \chi_{(-1,1)} + \frac{1}{2}\chi_{\{-1,1\}}$, and that $f_n$ (non-negative) is dominated by $f$ (which is Lebesgue integrable) for all $x,n$, you can use the dominated convergence theorem to argue that $$\int_{\mathbb{R}} f_n \xrightarrow[n\to\infty]{} \int_{\mathbb{R}} f = 2\ .$$

Clement C.
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$\quad$ All integrals of the form $~\displaystyle\int_0^\infty\dfrac{x^{k-1}}{(x^n+a^n)^m}~dx~$ can be evaluated by substituting $x=at$ and $u=\dfrac1{t^n+1}$ , then recognizing the expression of the beta function in the new integral, and lastly

employing Euler's reflection formula for the $\Gamma$ function to simplify the result.

Lucian
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