We model the density of tissues by an unknown function $f$ zero outside the zone to be studied. Assuming that each incident X-ray beam is carried by an affine line $\Delta$, and denoting by $I_e$ and $I_s$ its intensity measured on either side of the targeted zone: $$\ln\left(\frac{I_e}{I_s}\right) = \int_\Delta f$$
Propose a rigorous definition of the right-hand side of this equation in the case where $\Delta = \Delta\left(q, \vec{u}_\theta\right)$.

We model the density of tissues by an unknown function $f$ zero outside the zone to be studied. Assuming that each incident X-ray beam is carried by an affine line $\Delta$, and denoting by $I_e$ and $I_s$ its intensity measured on either side of the targeted zone: $$\ln\left(\frac{I_e}{I_s}\right) = \int_\Delta f$$
Explain how the Radon inversion formula allows us in principle to know the density of tissues in the radiographed zone.

For every natural number $n$, we denote by $S_{n}$ the function defined on $\mathbb{R}$ by
$$\forall x \in \mathbb{R}, \quad S_{n}(x) = \sum_{k=-n}^{n} e^{2\pi\mathrm{i} kx}$$
Let $f : \mathbb{R} \rightarrow \mathbb{C}$ be a function of class $C^{\infty}$ on $\mathbb{R}$ and 1-periodic. The sequence of complex numbers $(c_{n}(f))_{n \in \mathbb{Z}}$ is defined by
$$\forall n \in \mathbb{Z}, \quad c_{n}(f) = \int_{-1/2}^{1/2} f(x) e^{-2\pi\mathrm{i} nx} \mathrm{d}x$$
Let $n \in \mathbb{N}$. Calculate the integral $\int_{-1/2}^{1/2} S_{n}(x) \mathrm{d}x$.

Let $F$ be the vector subspace of $E$ formed of polynomial functions. For $k \in \mathbb{N}$, we denote by $p_k$ the function defined by $p_k(x) = x^k$. For all $k \in \mathbb{N}$, calculate $T(p_k)$. Deduce that $F$ is stable under $T$.

For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t)\,\mathrm{d}t$$ where $k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$ Let $f \in E$. Calculate $T(f)(0)$ and $T(f)(1)$.

Let $E_1$ denote the vector space of functions $f:[0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$. We set, for all $f \in E_1$, $$U(f)(s) = \int_0^1 k_s'(t) f'(t)\,\mathrm{d}t$$ where $k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$ Let $f \in E_1$ of class $\mathcal{C}^2$. Show that $U(f) = -T(f'')$. Deduce that $U(f) = f$.

Let $E_1$ denote the vector space of functions $f:[0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$. We set, for all $f \in E_1$, $$U(f)(s) = \int_0^1 k_s'(t) f'(t)\,\mathrm{d}t$$ Show that $U$ is the identity map on $E_1$.

We are given a real $a > 0$. We consider the space $E_3$ of functions $f:[0,a] \rightarrow \mathbb{R}$, continuous and of class $\mathcal{C}^1$ piecewise on $[0,a]$, and satisfying $f(0) = 0$. We equip $E_3$ with the inner product defined, for $f, g \in E_3$, by $$(f \mid g) = \int_0^a f'(t) g'(t)\,\mathrm{d}t$$ Show that the function $(x,y) \mapsto \min(x,y)$ is a reproducing kernel on $(E_3, (\cdot \mid \cdot))$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ Let $F$ be the vector subspace of $E$ formed by polynomial functions. For $k \in \mathbb{N}$, we denote by $p_k$ the function defined by $p_k(x) = x^k$. For all $k \in \mathbb{N}$, calculate $T(p_k)$. Deduce that $F$ is stable under $T$.

In this part, $E$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, equipped with the inner product defined by, $$\forall (f,g) \in E^2, \quad \langle f, g \rangle = \int_0^1 f(t) g(t) \, \mathrm{d}t$$ For all $s \in [0,1]$, we define the function $k_s$ by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ For all $f \in E$, we set, $$\forall s \in [0,1], \quad T(f)(s) = \int_0^1 k_s(t) f(t) \, \mathrm{d}t$$ Let $f \in E$. Calculate $T(f)(0)$ and $T(f)(1)$.

In this part, $E_1$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$. We denote by $N$ the norm associated with the inner product $(f \mid g) = \int_0^1 f'(t) g'(t) \, \mathrm{d}t$. For all $s \in [0,1]$, the function $k_s$ is defined by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ We set, for all $f \in E_1$, $$U(f)(s) = \int_0^1 k_s'(t) f'(t) \, \mathrm{d}t$$ Let $f \in E_1$ of class $\mathcal{C}^2$. Show that $U(f) = -T(f'')$. Deduce that $U(f) = f$.

In this part, $E_1$ denotes the vector space of functions $f : [0,1] \rightarrow \mathbb{R}$ continuous, of class $\mathcal{C}^1$ piecewise, and satisfying $f(0) = f(1) = 0$. We denote by $N$ the norm associated with the inner product $(f \mid g) = \int_0^1 f'(t) g'(t) \, \mathrm{d}t$. For all $s \in [0,1]$, the function $k_s$ is defined by, $$\forall t \in [0,1], \quad k_s(t) = \begin{cases} t(1-s) & \text{if } t < s \\ s(1-t) & \text{if } t \geqslant s. \end{cases}$$ We set, for all $f \in E_1$, $$U(f)(s) = \int_0^1 k_s'(t) f'(t) \, \mathrm{d}t$$ Show that $U$ is the identity map on $E_1$.

We are given a real $a > 0$. We consider the space $E_3$ of functions $f : [0,a] \rightarrow \mathbb{R}$, continuous and of class $\mathcal{C}^1$ piecewise on $[0,a]$, and satisfying $f(0) = 0$. We equip $E_3$ with the inner product defined, for $f, g \in E_3$, by $$(f \mid g) = \int_0^a f'(t) g'(t) \, \mathrm{d}t$$ Show that the function $(x,y) \mapsto \min(x,y)$ is a reproducing kernel on $(E_3, (\cdot \mid \cdot))$.

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ For $k \in \mathbb{N}$, what is the value of $m_{2k+1}$?

For every natural integer $k$ we set $$m_{k} = \frac{1}{2\pi} \int_{-2}^{2} x^{k} \sqrt{4 - x^{2}} \, \mathrm{d}x$$ For $k \in \mathbb{N}$, what is the value of $m_{2k+1}$?

Let $a , b , c , d$ be four real numbers such that $a \leqslant b$ and $c \leqslant d$. Let $U$ be an open set of $\mathbb { R } ^ { 2 }$ containing $[ a , b ] \times [ c , d ]$. Let $h : U \rightarrow \mathbb { R }$ be a function of class $\mathcal { C } ^ { 2 }$.
(a) Show the identity $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = \int _ { a } ^ { b } \hat { h } \left( s _ { 1 } \right) d s _ { 1 }$$ where $\hat { h }$ is defined by $$\hat { h } \left( s _ { 1 } \right) = \int _ { c } ^ { d } \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( s _ { 1 } , s _ { 2 } \right) d s _ { 2 }$$ (b) Deduce that there exists a point $\left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$ of $[ a , b ] \times [ c , d ]$ such that we have the two equalities $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = ( b - a ) \hat { h } \left( \bar { s } _ { 1 } \right) = ( b - a ) ( d - c ) \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$$

We keep, until the end of this third part, the hypotheses and notation of question 3.2. For $x , y \in I$ such that $y \neq x$, we set $$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$ (a) Show that for all $x , y \in I$ such that $y \neq x$ we have $$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$ (b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$.
(c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$.
(d) Compute $H _ { f } ( x , x )$.

We keep the hypotheses and notation of question 3.2. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$.
(a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that $$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$ (b) Compute $$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$ as a function of the derivatives of $f$.

Let $a , b , c , d$ be four real numbers such that $a \leqslant b$ and $c \leqslant d$. Let $U$ be an open set of $\mathbb { R } ^ { 2 }$ containing $[ a , b ] \times [ c , d ]$. Let $h : U \rightarrow \mathbb { R }$ be a function of class $\mathcal { C } ^ { 2 }$.
(a) Show the identity $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = \int _ { a } ^ { b } \hat { h } \left( s _ { 1 } \right) d s _ { 1 }$$ where $\hat { h }$ is defined by $$\hat { h } \left( s _ { 1 } \right) = \int _ { c } ^ { d } \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( s _ { 1 } , s _ { 2 } \right) d s _ { 2 }$$
(b) Deduce that there exists a point $\left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$ of $[ a , b ] \times [ c , d ]$ such that we have the two equalities $$h ( b , d ) - h ( a , d ) - h ( b , c ) + h ( a , c ) = ( b - a ) \hat { h } \left( \bar { s } _ { 1 } \right) = ( b - a ) ( d - c ) \frac { \partial ^ { 2 } h } { \partial s _ { 1 } \partial s _ { 2 } } \left( \bar { s } _ { 1 } , \bar { s } _ { 2 } \right)$$

We keep, until the end of this third part, the hypotheses and notation of the previous question. For $x , y \in I$ such that $y \neq x$, we set $$H _ { f } ( x , y ) = \frac { x f ( y ) - y f ( x ) } { f ( y ) - f ( x ) }$$
(a) Show that for all $x , y \in I$ such that $y \neq x$ we have $$H _ { f } ( x , y ) = x - f ( x ) \int _ { 0 } ^ { 1 } g ^ { \prime } ( \lambda f ( x ) + ( 1 - \lambda ) f ( y ) ) d \lambda$$
(b) Deduce that $H _ { f }$ admits a unique continuous extension to $I \times I$ as a whole. We still denote this extension by $H _ { f } : I \times I \rightarrow \mathbb { R }$.
(c) Show that $H _ { f }$ is of class $\mathcal { C } ^ { 2 }$ on $I \times I$.
(d) Compute $H _ { f } ( x , x )$.

We keep the hypotheses and notation of questions 3.2 and 3.3. We now assume $0 \in f ( I )$ and we denote $x ^ { * } = g ( 0 )$. For $x \in I$ we denote by $I _ { x }$ the closed interval with endpoints $x$ and $x ^ { * }$.
(a) Let $x , y \in I$. Show that there exists $( \bar { x } , \bar { y } ) \in I _ { x } \times I _ { y }$, such that $$H _ { f } ( x , y ) - x ^ { * } = \left( x - x ^ { * } \right) \left( y - x ^ { * } \right) \frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } ( \bar { x } , \bar { y } )$$
(b) Compute $$\frac { \partial ^ { 2 } H _ { f } } { \partial x \partial y } \left( x ^ { * } , x ^ { * } \right)$$ as a function of the derivatives of $f$.

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$ and set $J _ { \alpha } = \int _ { 0 } ^ { 1 } h _ { \alpha } ( t ) \, \mathrm { d } t$. Justify that $$J _ { \alpha } = \int _ { 0 } ^ { 1 } \ln ( 1 - t ) \, \mathrm { d } t + \int _ { 0 } ^ { 1 } \ln ( 1 + t ) \, \mathrm { d } t - \int _ { 0 } ^ { 1 } \ln \left( \alpha ^ { 2 } + t ^ { 2 } \right) \mathrm { d } t = \int _ { 0 } ^ { 2 } \ln ( u ) \, \mathrm { d } u - \int _ { 0 } ^ { 1 } \ln \left( \alpha ^ { 2 } + t ^ { 2 } \right) \mathrm { d } t.$$

For any real $\alpha > 0$, consider the function $h _ { \alpha } : t \mapsto \ln \left( \frac { 1 - t ^ { 2 } } { \alpha ^ { 2 } + t ^ { 2 } } \right)$ and set $J _ { \alpha } = \int _ { 0 } ^ { 1 } h _ { \alpha } ( t ) \, \mathrm { d } t$. Deduce that $$J _ { \alpha } = 2 \ln ( 2 ) - \ln \left( 1 + \alpha ^ { 2 } \right) - 2 \alpha \arctan \left( \frac { 1 } { \alpha } \right).$$

To each function $f \in E$, we associate the function $U ( f )$ 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 endomorphism $U$ surjective?

We consider two strictly positive real numbers $a$ and $b$, and we set $\rho = \frac{b-a}{b+a}$. We call $\Psi$ the application from $\mathbf{R}$ to $\mathbf{R}$ defined by: $$\forall x \in \mathbf{R}, \Psi(x) = \ln(a^2 \cos^2 x + b^2 \sin^2 x)$$ and $\sigma(x) = \sum_{k=1}^{+\infty} \frac{x^k}{k^2}$.
Conclude that $$\int_0^{\pi} \Psi(x)^2 \mathrm{d}x = 4\pi\left(\ln\left(\frac{a+b}{2}\right)\right)^2 + 2\pi\sigma(\rho^2)$$