<s>[INST] $I_n= \int_{0}^{ + \infty} \dfrac{ x^{4n-2} }{ ( 1+x^4)^n}$ 
*

*$x \geq 0$

*$n \geq 0$

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

*$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 $I_n \to 0$ ?
 [/INST] 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,
$$
I_{n + 1}  = I_1 \prod\limits_{k = 1}^n {\left( {1 - \frac{1}{{4k}}} \right)}  \to 0.
$$
</s>