Is Beta function essential to evaluating the integral $\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^n}$?

Question

After investigating the indefinite integral in the post, I found that

$$\int \frac { d x } { x ^ { 4 } + 1 } = \frac { 1 } { 4 \sqrt { 2 } } \left[ 2 \tan ^ { - 1 } \left( \frac { x ^ { 2 } - 1 } { \sqrt { 2 } x } \right) + \ln \left| \frac { x ^ { 2 } + \sqrt { 2 } x + 1 } { x ^ { 2 } - \sqrt { 2 } x + 1 } \right|\right] + C;$$

$$\int \frac{d x}{\left(1+x^{4}\right)^{2}}=\frac{x}{4\left(1+x^{4}\right)}+\frac{3}{16 \sqrt{2}}\left[2 \tan ^{-1}\left(\frac{x^{2}-1}{\sqrt{2} x}\right)+\ln \left|\frac{x^{2}+\sqrt{2} x+1}{x^{2}-\sqrt{2} x+1}\right|\right]+C $$ Plugging limits yields $$I_1= \int_0^{\infty} \frac{d x}{1+x^4}=\frac \pi{2\sqrt2};\quad I_2=\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^2}=\frac{3 \pi}{8 \sqrt{2}} $$ I guess about the value of the general integral $$I_n=\int_0^{\infty} \frac{d x}{\left(1+x^4\right)^n} =\frac {p\pi}{q\sqrt 2} $$ The form of the integrand reminds me that Beta function may help. Then I put the substitution $y=x^4$ and get $$ I_n =\frac{1}{4} \int_0^{\infty} \frac{y^{-\frac{3}{4}} d y}{(1+y)^n} =\frac{1}{4} B\left(\frac{1}{4}, n-\frac{1}{4}\right) $$ Expressing in terms of Gamma function gives $$ \begin{aligned}I_n&=\frac{1}{4} \frac{\Gamma\left(\frac{1}{4}\right) \Gamma\left(n-\frac{1}{4}\right)}{\Gamma(n)} \\&= \frac{1}{4(n-1) !} \Gamma\left(\frac{1}{4}\right) \Gamma\left(n-\frac{1}{4}\right)\\ &=\frac{1}{4(n-1)!} \Gamma\left(\frac{1}{4}\right)\left(n-1-\frac{1}{4}\right)\left(n-2-\frac{1}{4}\right) \cdots \frac{3}{4} \Gamma\left(\frac{3}{4}\right) \cdots (*) \\ &=\frac{1}{4(n-1) !} \left[\frac{1}{4^{n-1}}(4 n-5)(4 n-9) \cdots 3 \right] \pi \csc \left(\frac{\pi}{4}\right) \cdots (**)\\ &=\frac{\pi \sqrt{2}}{4^n(n-1) !} \prod_{k=1}^{n-1}(4 k-1) \end{aligned} $$ (*) uses $\Gamma(z+1)=z \Gamma(z)$ for all $z\in \mathbb{C}$ and

(**) uses $\Gamma(x) \Gamma(1-x)=\pi \csc (\pi x)$ for $z\not \in \mathbb{Z}.$

For examples, $$ \begin{aligned} & I_3=\frac{\pi \sqrt{2}}{128} \cdot 3 \cdot 7=\frac{21 \pi \sqrt{2}}{128} \\ & I_4=\frac{\pi \sqrt{2}}{256 \cdot 6} \cdot 3 \cdot 7 \cdot 11=\frac{77 \pi \sqrt{2}}{512}\\& I_{10}=\frac{\pi \sqrt{2}}{4^{10} \cdot 9 !} \prod_{k=1}^{9}(4 k-1)=\frac{15646785 \pi \sqrt{2}}{134217728} \end{aligned} $$ checked by WA.

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My Question: Are there any alternative methods?

Your comments and alternative solutions are highly appreciated.

Answer 1

Any standard method (including scaling $x$ in the first display equation in the question statement) yields $$\int_0^\infty \frac{dx}{a + x^4} = \frac{\pi}{2 \sqrt 2} a^{-3 / 4} .$$ If $n$ is a positive integer, differentiating gives \begin{align} \frac{d^{n - 1}}{da^{n - 1}} \int_0^\infty \frac{dx}{a + x^4} &= \frac{d^{n - 1}}{da^{n - 1}} \left(\frac{\pi}{2 \sqrt 2} a^{-3 / 4} \right) \\ (-1)^n (n - 1)!\int_0^\infty \frac{dx}{(a + x^4)^n} &= \frac{\pi}{2 \sqrt 2} \underbrace{\left(-\frac{3}{4}\right)\left(-\frac{7}{4}\right) \cdots \left(\frac54 - n \right)}_{n - 1} a^{\frac{1}{4} - n} . \end{align} Evaluating at $a = 1$ and rearranging gives $$\int_0^\infty \frac{dx}{(1 + x^4)^n} = \frac{\pi \sqrt 2}{4^n (n - 1)!} \prod_{k = 1}^{n - 1} (4 k - 1),$$ which agrees with your solution.

Answer 2

\begin{align}n\geq 1,J_n=\int_0^\infty \frac{1}{(1+x^4)^n}dx\end{align} For $0\leq t< 1$, Define, \begin{align}F(t)&=\sum_{n=1}^\infty J_nt^n\\ &=\sum_{n=1}^\infty\left(\int_0^\infty \frac{t^n}{(1+x^4)^n}dx\right)\\ &=\int_0^\infty \left(\sum_{n=1}^\infty \left(\frac{t}{1+x^4}\right)^n\right)dx\\ &=\int_0^\infty \frac{\frac{t}{1+x^4}}{1-\frac{t}{1+x^4}}dx\\ &=\int_0^\infty \frac{t}{1-t+x^4}dx\\ &\overset{u=\frac{x}{\left(1-t\right)^{\frac{1}{4}}}}=\frac{t}{(1-t)^{\frac{3}{4}}}\underbrace{\int_0^\infty\frac{1}{1+u^4}du}_{=K}\\ \end{align}

\begin{align}K&=\overset{z=\frac{1}{u}}=\int_0^\infty\frac{z^2}{1+z^4}dz\\ 2K&=\int_0^\infty \frac{1+z^2}{1+z^4}dz=\int_0^\infty \frac{1+\frac{1}{z^2}}{\left(z-\frac{1}{z}\right)^+2}dz\\ &\overset{w=z-\frac{1}{z}}=\int_{-\infty}^{+\infty}\frac{1}{2+w^2}dw=\frac{1}{\sqrt{2}}\left[\arctan\left(\frac{w}{\sqrt{2}}\right)\right]_{-\infty}^{+\infty}\\ &\boxed{K=\frac{\pi}{2\sqrt{2}},F(t)=\frac{\pi t}{2(1-t)^{\frac{3}{4}}\sqrt{2}}}\ \end{align} Thus, \begin{align}n\geq 1,J_n&=\frac{F^{(n)}(0)}{n!}\\ &=\frac{\pi}{n!\times 2\sqrt{2}}\sum_{k=0}^{n}\binom{n}{k}(t)^{(k)}(0)\left((1-t)^{-\frac{3}{4}}\right)^{(n-k)}(0)\\ &=\frac{\binom{n}{1}\pi\left((1-t)^{-\frac{3}{4}}\right)^{(n-1)}(0)}{n!\times 2\sqrt{2}}=\frac{\pi\left((1-t)^{-\frac{3}{4}}\right)^{(n-1)}(0)}{(n-1)!\times 2\sqrt{2}} \end{align}

Moreover, \begin{align}\big((1-x)^r\Big)^{(n)}=(-1)^{n}r(r-1)...\Big(r-(n-1)\Big)(1-x)^{r-n}\end{align}

Therefore, \begin{align}\boxed{J_n=\frac{\pi\prod_{k=0}^{n-2}(3+4k)}{{2^{2(n-1)}(n-1)!\times 2\sqrt{2}}}}\end{align}

PS: Or,
\begin{align}\boxed{J_n=\frac{\pi\sqrt{2}\prod_{k=1}^{n-1}(4k-1)}{{2^{2n}(n-1)!}}}\end{align}

Answer 3

The antiderivative $$I=\int \frac {dx}{ \left(1+x^a\right)^n}=x \,\, _2F_1\left(\frac{1}{a},n;1+\frac{1}{a};-x^a\right)$$ So, the definite integral is $$J=\int_0^\infty \frac {dx}{ \left(1+x^a\right)^n}=\frac{\Gamma \left(1+\frac{1}{a}\right) \Gamma \left(n-\frac{1}{a}\right)}{\Gamma (n)}\quad \text{if}\quad a n>1$$

Answer 4

We are glad to have 3 nice and inspiring alternative solutions. I now want to give one more by introducing a reduction formula : $$ I_{n+1}=\frac{4 n-1}{4 n} I_n \text { for any } n\in \mathbb{N}. $$ Using integration by parts, we have $$ \begin{aligned} J_n & =\int \frac{1}{\left(1+x^4\right)^n} d x \\ & =-\frac{1}{4(n-1)} \int \frac{1}{x^3} d\left[\frac{1}{\left(1+x^4\right)^{n-1}}\right] \\ & =-\frac{1}{4(n-1)}\left[\frac{1}{x^3\left(1+x^4\right)^{n-1}}+3 \int \frac{1}{x^4\left(1+x^4\right)^{n-1}} d x\right] \\& =-\frac{1}{4(n-1)}\left[\frac{1}{x^3\left(1+x^4\right)^{n-1}}+3 \int \left(\frac{1}{x^4}-\frac{1}{1+x^4}-\frac{1}{\left(1+x^4\right)^2}-\cdots-\frac{1}{\left(1+x^4\right)^{n-1}}\right)dx\right]\\ & =-\frac{1}{4(n-1)} \left[\frac{1-\left(1+x^4\right)^{n-1}}{x^3\left(1+x^4\right)^{n-1}}\right]+\frac{3}{4 n-1}\left(J_1+J_2+\ldots+J_{n-1}\right) \end{aligned} $$ Taking limits yields $$ I_n=\frac{3}{4(n-1)}\left(I_1+I_2+\ldots+I_{n-1}\right) $$$$ \begin{aligned} & 4(n-1) I_n=3\left(I_1+I_2+\ldots+I_{n-1}\right) \\ & 4(n-1) I_n+3 I_n=3\left(I_1+I_2+\ldots+I_n\right) \\ & (4 n-1) I_n=4 n I_{n+1} \end{aligned} $$ So we get$$I_{n+1}=\frac{4 n-1}{4 n} I_n \text { for any } n\in \mathbb{N} $$

Hence $$ \begin{aligned} I_n & =\frac{4 n-5}{4(n-1)} I_{n-1} \\ & =\frac{4 n-5}{4(n-1)} \cdot \frac{4 n-9}{4(n-2)} \cdots \frac{3}{4} I_1 \\ & =\frac{1}{4^{n-1}(n-1) !} \prod_{k=1}^{n-1}(4 k-1) \cdot \frac{\pi}{2 \sqrt{2}} \\ & =\frac{\pi \sqrt{2}}{4^n(n-1) !} \prod_{k=1}^{n-1}(4 k-1) \end{aligned} $$

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After submitting the solution, I found that the reduction formula can be easily found by integration by parts on $1dx$ as: $$ \begin{aligned} I_n & =\int_0^{\infty} \frac{1}{\left(1+x^4\right)^n} d x \\ & =\left[\frac{x}{\left(1+x^4\right)^n}\right]_0^{\infty}+n \int_0^{\infty} x \cdot\frac{4 x^3}{\left(1+x^4\right)^{n+1}} d x \\ & =4 n \int_0^{\infty} \frac{x^4+1-1}{\left(1+x^4\right)^{n+1}} d x \\ & =4 n\left[\int_0^{\infty} \frac{1}{\left(1+x^4\right)^n} d x-\int_0^{\infty} \frac{1}{\left(1+x^2\right)^{n+1}} d x\right] \\ & =4 n\left[I_n-I_{n+1}\right] \end{aligned} $$ Rearranging gives $$I_{n+1}=\frac{4 n-1}{4 n} I_n \text { for any } n\in \mathbb{N} $$