$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$ 
*

*$x \geq 0$

*$n \geq 0$

*$f_n(x)= \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$

*$I_n= \int_{0}^{ + \infty}f_n(x) $

*What is the limit of $I_n$ ?

*What is an equivalent of $I_n$ ?


$
\begin{align*}
u'&= (-n+1) 4x^3 (1+x^4)^{-n-1}\\
u &=x^{4n-1}\\
v &= x^{4n-1} \\
v' &= (4n-1) x^{4n-2} \\
I_{n+1}&=  \int_{0}^{ + \infty} \dfrac{-4n x^3 x^{4n-1}   }{ 4 (1+x^4)^{n+1} (-n) }\\
I_{n+1}&= \left[ \dfrac{ x^{4n-1}}{-4n (1+x^4)^n }  \right] +   \int_{0}^{ + \infty} \dfrac{(4n-1) x^{4n-2}}{4n (1+x^4)^n } \\
I_{n+1} &= \dfrac{4n-1}{4n}I_n \\
I_2 &=\dfrac{4-1}{4 \times 1}I_1 \\
I_{n+1} &= \dfrac{\prod_{k=1}^{n} (4k-1) }{4^n n!} I_1  \\
I_1&=  \int_{0}^{ + \infty} \dfrac{ x^{2}}{ (1+x^4) } dx \\
\end{align*}
$
Quanto's answer for $I_1$ :
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx 
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx = \frac12 \int_0^{\infty} \dfrac{ 1+\frac1{x^2}}{ x^2+\frac1{x^2} } dx \\
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
$I_{n+1} =  \dfrac{ \prod_{k=1}^n (4k-1) }{ 4^n n!} I_1 =\prod_{k=1}^n \dfrac{4k-1}{4k} I_1=\prod_{k=1}^n (1- \dfrac{1}{4k}) I_1$
What is the limit ?
$\ln( 1 - u) \sim u$ donc $\ln (1 - \dfrac{1}{4k}) \sim -  \dfrac{1}{4k}$
and $- \sum \dfrac{1}{k} = - \infty$ so $I_n \to 0$ ?
 A: Complex analysis is one possibility. Note that $I_1 = \frac{1}{2} \int_{-\infty}^{\infty} \frac{x^2}{1+x^4} \ dx = J$. Consider the semi-circle contour $C$ in the upper half of the complex plane and note that there are two poles (at $z_1 = e^{\pi i/4}$ and $z_2 = e^{3\pi i/4}$):
$$ 2\pi i \left( \frac{1}{4z_1} + \frac{1}{4z_2} \right) = \oint_C \frac{z^2}{1 + z^4} \ dz = J + \int_{C_R} \frac{z^2}{1+z^4} \ dz $$
where $C_R: z = R e^{i \theta}$ with $\theta \in [0,\pi]$. You can show that $|\int_{C_R}| \to 0$ as $R \to \infty$ and simplifying the above expression you obtain
$$ J = \frac{\pi}{\sqrt 2} \implies I_1 = \frac{\pi}{2\sqrt 2} $$
A: Proceed as follows
$$ \int_0^{\infty} \dfrac{ x^{2}}{ 1+x^4 } dx 
\overset{x\to \frac1x}=\frac12\int_0^{\infty} \dfrac{ 1+x^2}{ 1+x^4 } dx 
= \frac12 \int_0^{\infty} \dfrac{ d(x-\frac1{x})}{ (x-\frac1{x})^2+2 } dx =\frac\pi{2\sqrt2}
$$
A: If you like the gaussian hypergeometric function
$$J_n= \int \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}\,dx=\frac{x^{4 n-1} }{4 n-1}\,\,_2F_1\left(n-\frac{1}{4},n;n+\frac{3}{4};-x^4\right)$$  Assuming that $n \gt \frac 14$, for $x=0$ the rhs tends to $0$ and
$$I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}=\frac{\Gamma \left(\frac{1}{4}\right) \Gamma \left(n+\frac{3}{4}\right)}{(4
   n-1) \Gamma (n)}=\frac{\Gamma \left(\frac{5}{4}\right) \Gamma \left(n-\frac{1}{4}\right)}{\Gamma(n)}$$ which the same as @Svyatoslav's result in comments.
Now, consider
$$y=\frac{\Gamma \left(n-\frac{1}{4}\right)}{\Gamma(n)}\implies \log(y)=\log \left(\Gamma \left(n-\frac{1}{4}\right)\right)-\log \left(\Gamma \left(n\right)\right)$$ Apply Stirling approximation twice and continue with Taylor expansion
$$\log(y)=\frac{1}{4} \log \left(\frac{1}{n}\right)-\frac{3}{32 n}+\frac{1}{128   n^2}+O\left(\frac{1}{n^3}\right)$$
$$y\sim\frac 1 {n^{\frac 14}} \exp\Big[-\frac{3}{32 n}+\frac{1}{128   n^2}\Big]$$ and then the approximation
$$I_n\sim\frac {\Gamma \left(\frac{5}{4}\right) } {n^{\frac 14}} \exp\Big[-\frac{3}{32 n}+\frac{1}{128   n^2}\Big]$$
Trying for $n=10^4$, the exact value is $0.0906417$ while the approximation gives $0.0906394$.
A: Note that
$$
\log (1 - x) <  - x
$$
for $0<x<1$. Therefore,
\begin{align*}
0 & < \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)}  = \exp \left( {\sum\limits_{k = 1}^n {\log \left( {1 - \frac{1}{{4k}}} \right)} } \right) \\ & < \exp \left( { - \sum\limits_{k = 1}^n {\frac{1}{{4k}}} } \right)  < \exp \left( { - \frac{1}{4}\log n} \right) = \frac{1}{{n^{1/4} }}.
\end{align*}
Thus, by the squeeze theorem,
$$
I_{n + 1}  = I_1 \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)}  \to 0.
$$
