Let $z \in D$. Show the convergence of the series $\sum _ { n \geq 1 } \frac { z ^ { n } } { n }$. Specify the value of its sum when $z \in ] - 1,1 [$. We denote $$L ( z ) : = \sum _ { n = 1 } ^ { + \infty } \frac { z ^ { n } } { n }$$
Q2
First order differential equations (integrating factor)View
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 $\Psi : t \mapsto ( 1 - t z ) e ^ { L ( t z ) }$ 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 $\sum _ { n \geq 1 } L \left( z ^ { n } \right)$ is convergent for all $z$ in $D$.
Q5
First order differential equations (integrating factor)View
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 $z \in D$.
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$. Show that $q$ is piecewise continuous on $\mathbf { R }$, that it is 1-periodic, and that the function $| q |$ is even.
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$. Show that $\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u$ is well-defined for all real $t > 0$.
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$. Show that for all integer $n > 1$, $$\int _ { 1 } ^ { n } \frac { q ( u ) } { u } \mathrm {~d} u = \ln ( n ! ) + ( n - 1 ) - n \ln ( n ) - \frac { 1 } { 2 } \ln ( n ) = \ln \left( \frac { n ! e ^ { n } } { n ^ { n } \sqrt { n } } \right) - 1$$
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$, where $\lfloor x \rfloor$ denotes the integer part of $x$. Show that $\int _ { \lfloor x \rfloor } ^ { x } \frac { q ( u ) } { u } \mathrm {~d} u$ tends to 0 as $x$ tends to $+ \infty$, and deduce the convergence of the integral $\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u$, as well as the equality $$\int _ { 1 } ^ { + \infty } \frac { q ( u ) } { u } \mathrm {~d} u = \frac { \ln ( 2 \pi ) } { 2 } - 1$$
Using a series expansion under the integral, show that $$\int _ { 0 } ^ { + \infty } \ln \left( 1 - e ^ { - u } \right) \mathrm { d } u = - \frac { \pi ^ { 2 } } { 6 }$$
Show that $$\int _ { 0 } ^ { 1 } \ln \left( \frac { 1 - e ^ { - t u } } { t } \right) \mathrm { d } u \underset { t \rightarrow 0 ^ { + } } { \longrightarrow } - 1$$ You may begin by establishing that $x \mapsto \frac { 1 - e ^ { - x } } { x }$ is decreasing on $\mathbf { R } _ { + } ^ { * }$.
For $k \in \mathbf { N } ^ { * }$ and $t \in \mathbf { R } _ { + }$, we set $$u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u \quad \text { if } t > 0 , \quad \text { and } \quad u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { q ( u ) } { u } \mathrm {~d} u \quad \text { if } t = 0$$ where $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$. Show that $u _ { k }$ is continuous on $\mathbf { R } _ { + }$ for all $k \in \mathbf { N } ^ { * }$.
For $k \in \mathbf { N } ^ { * }$ and $t \in \mathbf { R } _ { + }$, we set $$u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u \quad \text { if } t > 0 , \quad \text { and } \quad u _ { k } ( t ) = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { q ( u ) } { u } \mathrm {~d} u \quad \text { if } t = 0$$ where $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$. Let $t \in \mathbf { R } _ { + } ^ { * }$. Show successively that $\left| u _ { k } ( t ) \right| = \int _ { k/2 } ^ { ( k + 1 ) / 2 } \frac { t | q ( u ) | } { e ^ { t u } - 1 } \mathrm {~d} u$, then $u _ { k } ( t ) = ( - 1 ) ^ { k } \left| u _ { k } ( t ) \right|$ for all integer $k \geq 1$, and finally establish that $$\forall n \in \mathbf { N } ^ { * } , \left| \sum _ { k = n } ^ { + \infty } u _ { k } ( t ) \right| \leq \frac { 1 } { 2 n } .$$ We admit in what follows that this bound also holds for $t = 0$.
The function $q$ associates to any real $x$ the real number $q ( x ) = x - \lfloor x \rfloor - \frac { 1 } { 2 }$. Show, for all real $t > 0$, the identity $$\int _ { 1 } ^ { + \infty } \frac { t q ( u ) } { e ^ { t u } - 1 } \mathrm {~d} u = - \frac { 1 } { 2 } \ln \left( 1 - e ^ { - t } \right) - \ln P \left( e ^ { - t } \right) - \int _ { 1 } ^ { + \infty } \ln \left( 1 - e ^ { - t u } \right) \mathrm { d } u$$
We have $P ( z ) = \sum _ { n = 0 } ^ { + \infty } p _ { n } z ^ { n }$ for all $z \in D$, where $p_n$ denotes the number of partitions of $n$. Let $n \in \mathbf { N }$. Show that for all real $t > 0$, $$p _ { n } = \frac { e ^ { n t } P \left( e ^ { - t } \right) } { 2 \pi } \int _ { - \pi } ^ { \pi } e ^ { - i n \theta } \frac { P \left( e ^ { - t } e ^ { i \theta } \right) } { P \left( e ^ { - t } \right) } \mathrm { d } \theta \tag{1}$$
Let $x \in [ 0,1 [$ and $\theta \in \mathbf { R }$. Using the function $L$, show that $$\left| \frac { 1 - x } { 1 - x e ^ { i \theta } } \right| \leq \exp ( - ( 1 - \cos \theta ) x ) .$$ Deduce that for all $x \in [ 0,1 [$ and all real $\theta$, $$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 1 - x } + \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \right)$$
Let $x \in [ 0,1 [$ and $\theta$ a real number. Show that $$\frac { 1 } { 1 - x } - \operatorname { Re } \left( \frac { 1 } { 1 - x e ^ { i \theta } } \right) \geq \frac { x ( 1 - \cos \theta ) } { ( 1 - x ) \left( ( 1 - x ) ^ { 2 } + 2 x ( 1 - \cos \theta ) \right) }$$ Deduce that if $x \geq \frac { 1 } { 2 }$ then $$\left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 - \cos \theta } { 6 ( 1 - x ) ^ { 3 } } \right) \quad \text { or } \quad \left| \frac { P \left( x e ^ { i \theta } \right) } { P ( x ) } \right| \leq \exp \left( - \frac { 1 } { 3 ( 1 - x ) } \right)$$ For this last result, distinguish two cases according to the relative values of $x ( 1 - \cos \theta )$ and $( 1 - x ) ^ { 2 }$.