Let $z \in D$. Show that the function $t \in [ 0,1 ] \mapsto L ( t z )$ is differentiable and give a simple expression for its derivative. Deduce that $t \mapsto ( 1 - t z ) e ^ { L ( t z ) }$ is constant on $[ 0,1 ]$ and conclude that
$$\exp ( L ( z ) ) = \frac { 1 } { 1 - z }$$

Let $z \in D$. Show that the function $\Phi : t \mapsto L ( t z )$ is differentiable on an open interval including $[ - 1,1 ]$ and give a simple expression for its derivative on $[ - 1,1 ]$.

Let $z \in D$. Show that the function $\Psi : t \mapsto (1-tz)e^{L(tz)}$ is constant on $[0,1]$, and deduce that $$\exp(L(z)) = \frac{1}{1-z}$$

Show that $|L(z)| \leq -\ln(1-|z|)$ for all $z$ in $D$. Deduce that the series $\sum_{n \geq 1} L(z^n)$ is convergent for all $z$ in $D$.

Let $z \in D$. Verify that $P ( z ) \neq 0$, that
$$P ( z ) = \lim _ { N \rightarrow + \infty } \prod _ { n = 1 } ^ { N } \frac { 1 } { 1 - z ^ { n } }$$
and that for all real $t > 0$,
$$\ln P \left( e ^ { - t } \right) = - \sum _ { n = 1 } ^ { + \infty } \ln \left( 1 - e ^ { - n t } \right)$$
where $P ( z ) := \exp \left[ \sum _ { n = 1 } ^ { + \infty } L \left( z ^ { n } \right) \right]$ for all $z \in D$.

We fix a real $\alpha > 0$ and an integer $n \geq 1$. Subject to existence, we set
$$S _ { n , \alpha } ( t ) : = \sum _ { k = 1 } ^ { + \infty } \frac { k ^ { n } e ^ { - k t \alpha } } { \left( 1 - e ^ { - k t } \right) ^ { n } }$$
We also introduce the function
$$\varphi _ { n , \alpha } : x \in \mathbf { R } _ { + } ^ { * } \mapsto \frac { x ^ { n } e ^ { - \alpha x } } { \left( 1 - e ^ { - x } \right) ^ { n } }$$
which is obviously of class $\mathcal { C } ^ { \infty }$.
Show that $\varphi _ { n , \alpha }$ and $\varphi _ { n , \alpha } ^ { \prime }$ are integrable on $] 0 , + \infty [$.

We fix a real $\alpha > 0$ and an integer $n \geq 1$. Subject to existence, we set
$$S _ { n , \alpha } ( t ) : = \sum _ { k = 1 } ^ { + \infty } \frac { k ^ { n } e ^ { - k t \alpha } } { \left( 1 - e ^ { - k t } \right) ^ { n } }$$
We also introduce the function
$$\varphi _ { n , \alpha } : x \in \mathbf { R } _ { + } ^ { * } \mapsto \frac { x ^ { n } e ^ { - \alpha x } } { \left( 1 - e ^ { - x } \right) ^ { n } }$$
Show, for all real $t > 0$, the existence of $S _ { n , \alpha } ( t )$, its strict positivity, and the identity
$$\int _ { 0 } ^ { + \infty } \varphi _ { n , \alpha } ( x ) \mathrm { d } x = t ^ { n + 1 } S _ { n , \alpha } ( t ) - \sum _ { k = 0 } ^ { + \infty } \int _ { k t } ^ { ( k + 1 ) t } ( x - k t ) \varphi _ { n , \alpha } ^ { \prime } ( x ) \mathrm { d } x$$
Deduce that
$$S _ { n , \alpha } ( t ) = \frac { 1 } { t ^ { n + 1 } } \int _ { 0 } ^ { + \infty } \frac { x ^ { n } e ^ { - \alpha x } } { \left( 1 - e ^ { - x } \right) ^ { n } } \mathrm {~d} x + O \left( \frac { 1 } { t ^ { n } } \right) \quad \text { as } t \rightarrow 0 ^ { + }$$

For all $n \in \mathbb{N}^{\star}$, justify that there exists a unique function $f_n \in \mathcal{C}^{2}(]0, +\infty[)$ satisfying $f_n(1) = 0$, $f_n(2) = 0$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$ for all $x > 0$.

Show that the function series $\sum f_n(x)$ converges normally on any segment contained in $]0, +\infty[$ and that the function $F : x \mapsto \sum_{n=1}^{+\infty} f_n(x)$ is of class $\mathcal{C}^2$ on $]0, +\infty[$.

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \left\langle k _ { x } \mid f \right\rangle = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Using the Cauchy-Schwarz inequality, show that for all functions $f \in E$, $$\lim _ { \substack { x \rightarrow 0 \\ x > 0 } } U ( f ) ( x ) = 0$$

Explicitly determine $F^{\prime\prime}(x)$, where $F(x) = \sum_{n=1}^{+\infty} f_n(x)$ and $f_n^{\prime\prime}(x) = (-1)^n 2^{-nx^2}$.

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \left\langle k _ { x } \mid f \right\rangle = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Show that for all functions $f \in E$ and for all $x > 0$, $$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$

To each function $f \in E$, we associate the function $U ( f )$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { x } \left( 1 - \mathrm { e } ^ { - t } \right) \frac { f ( t ) } { t } \mathrm {~d} t + \left( \mathrm { e } ^ { x } - 1 \right) \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Let $f \in E$. Show that $U ( f )$ is of class $\mathcal { C } ^ { 1 }$ on $\mathbb { R } _ { + } ^ { * }$ and satisfies, for all $x > 0$, $$( U ( f ) ) ^ { \prime } ( x ) = \mathrm { e } ^ { x } \int _ { x } ^ { + \infty } f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$

For $p \in \mathbb { N } ^ { * }$, let $h(x) = \mathrm{e}^{-x} P(x)$ where $P$ is a polynomial solution of $(E_p)$. We denote by $\left( b _ { n } \right) _ { n \in \mathbb { N } }$ the sequence of coefficients of the power series expansion of $h$, so that for all $x \in \mathbb { R }$, $h ( x ) = \sum _ { n = 0 } ^ { + \infty } b _ { n } x ^ { n }$. These coefficients satisfy $$\left\{ \begin{array} { l } b _ { 0 } = 0 \\ n ( n + 1 ) b _ { n + 1 } = - ( n + p ) b _ { n } , \quad \forall n \in \mathbb { N } ^ { * } . \end{array} \right.$$ Establish that, for all $n \in \mathbb { N } ^ { * } , b _ { n } = \frac { ( - 1 ) ^ { n - 1 } ( n + p - 1 ) ! } { p ! n ! ( n - 1 ) ! } b _ { 1 }$.

For $p \in \mathbb { N } ^ { * }$, let $h(x) = \mathrm{e}^{-x} P(x)$ where $P$ is a polynomial solution of $(E_p)$, with power series coefficients satisfying $b _ { n } = \frac { ( - 1 ) ^ { n - 1 } ( n + p - 1 ) ! } { p ! n ! ( n - 1 ) ! } b _ { 1 }$ for all $n \in \mathbb{N}^*$. We set $g _ { p } ( x ) = x ^ { p - 1 } \mathrm { e } ^ { - x }$. Justify that $g _ { p } ^ { ( p ) }$ is developable as a power series and deduce from Question 34 that, for all $x \in \mathbb { R }$, $$P ( x ) = C x \mathrm { e } ^ { x } g _ { p } ^ { ( p ) } ( x )$$ where $C$ is a real constant whose expression in terms of $b _ { 1 }$ and $p$ we will specify.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p) : x(y'' - y') + py = 0$, with coefficients satisfying $n(n+1)a_{n+1} = (n-p)a_n$ for all $n \in \mathbb{N}^*$. Show that there exists a natural integer $q > p$ such that, for all integer $n \geqslant q$, $$\left| a _ { n + 1 } \right| \geqslant \frac { \left| a _ { n } \right| } { 2 ( n + 1 ) }.$$

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$, with $q$ a natural integer such that $q > p$ and $\left| a _ { n + 1 } \right| \geqslant \frac { \left| a _ { n } \right| } { 2 ( n + 1 ) }$ for all $n \geqslant q$. Deduce that, for all integer $n \geqslant q , \left| a _ { n } \right| \geqslant \frac { q ! \left| a _ { q } \right| } { 2 ^ { n - q } n ! }$.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$, with $q$ a natural integer such that $q > p$ and $\left| a _ { n } \right| \geqslant \frac { q ! \left| a _ { q } \right| } { 2 ^ { n - q } n ! }$ for all $n \geqslant q$. Show that the function $\psi : \left\lvert\, \begin{array} { c c c } \mathbb { R } _ { + } ^ { * } & \rightarrow & \mathbb { R } \\ x & \mapsto & \sum _ { n = q } ^ { + \infty } \left| a _ { n } \right| x ^ { n } \end{array} \right.$ is not an element of $E$.

For $p \notin \mathbb{N}^*$, $p \neq 0$, let $f(x) = \sum_{n=0}^{+\infty} a_n x^n$ be a non-zero power series solution of $(E_p)$. Using the result of Question 39, deduce finally that the function $f$ is not an element of $E$.

To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ Is the real number $0$ an eigenvalue of $U$?

To each function $f \in E$, we associate the endomorphism $U$ of $E$ defined for all $x > 0$ by $$U ( f ) ( x ) = \int _ { 0 } ^ { + \infty } \left( \mathrm { e } ^ { \min ( x , t ) } - 1 \right) f ( t ) \frac { \mathrm { e } ^ { - t } } { t } \mathrm {~d} t$$ It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ on $\mathbb{R}_+^*$. Let $\lambda \in \mathbb { R } ^ { * }$. We assume that $\lambda$ is an eigenvalue of $U$. Let $f$ be an eigenvector associated with it. Show that $f$ is a solution of the differential equation $(E_{1/\lambda}) : x(y'' - y') + \frac{1}{\lambda} y = 0$.

To each function $f \in E$, we associate the endomorphism $U$ of $E$. Let $\lambda \in \mathbb{R}^*$ be an eigenvalue of $U$ with eigenvector $f$, which is a solution of $(E_{1/\lambda})$. We assume that $f$ is developable as a power series on $\mathbb { R } _ { + } ^ { * }$, that is, there exists a power series $\sum _ { n \geqslant 0 } a _ { n } x ^ { n }$ of infinite radius of convergence such that $$\forall x \in \mathbb { R } _ { + } ^ { * } , \quad f ( x ) = \sum _ { n = 0 } ^ { + \infty } a _ { n } x ^ { n } .$$ Using the results of Part IV, show that the only possible eigenvalues of $U$ are of the form $\lambda = 1 / p$ with $p \in \mathbb { N } ^ { * }$.

For $p \in \mathbb{N}^*$, let $P$ be a non-zero polynomial solution of $(E_p) : x(y'' - y') + py = 0$. It has been shown that $U(f)$ satisfies $y'' - y' = -f(x)/x$ and that $U$ is self-adjoint ($\langle f | U(g) \rangle = \langle U(f) | g \rangle$). Prove that the function $pU(P) - P$ satisfies on $\mathbb { R } _ { + } ^ { * }$ the differential equation $y ^ { \prime \prime } - y ^ { \prime } = 0$.

For $p \in \mathbb{N}^*$, let $P$ be a non-zero polynomial solution of $(E_p) : x(y'' - y') + py = 0$. It has been shown that $pU(P) - P$ satisfies $y'' - y' = 0$ on $\mathbb{R}_+^*$. Show that $P$ is an eigenvector of $U$ for the eigenvalue $1/p$.

For $0 < \mu \leqslant 1$, we consider $F_\mu$ defined by: $$\forall y \in ]0, +\infty[, \quad F_\mu(y) = \frac{a}{\mu} y\left(1 - \left(\frac{y}{\theta}\right)^\mu\right)$$ with $a, \theta > 0$ and $0 < y_{\text{init}} < \theta$. By considering the function $z_\mu(t) = \phi_\mu(t)^{-\mu}$ find the expression of the solution $\phi_\mu$ on $[0, +\infty[$ associated with $F_\mu$.