Let $f$ be a function such that $\int _ { 6 } ^ { 12 } f ( 2 x ) d x = 10$. Which of the following must be true?
(A) $\int _ { 12 } ^ { 24 } f ( t ) d t = 5$
(B) $\int _ { 12 } ^ { 24 } f ( t ) d t = 20$
(C) $\int _ { 6 } ^ { 12 } f ( t ) d t = 5$
(D) $\int _ { 6 } ^ { 12 } f ( t ) d t = 20$
(E) $\int _ { 3 } ^ { 6 } f ( t ) d t = 5$

Let $\varepsilon$ and $r$ be fixed such that $0 < \varepsilon < r$. With the change of variables $q = r\cos\theta$, establish that $$\int_\varepsilon^r \frac{\mathrm{d}q}{q^2\sqrt{r^2-q^2}} = \frac{\sqrt{r^2-\varepsilon^2}}{r^2\varepsilon}$$

For $( x , y )$ in $\left( \mathbb { R } ^ { + * } \right) ^ { 2 }$, we define $\beta ( x , y ) = \int _ { 0 } ^ { 1 } t ^ { x - 1 } ( 1 - t ) ^ { y - 1 } \mathrm {~d} t$. Throughout the rest of this question we assume $x > 1$ and $y > 1$.
Show that $\beta ( x , y ) = \int _ { 0 } ^ { + \infty } \frac { u ^ { x - 1 } } { ( 1 + u ) ^ { x + y } } \mathrm {~d} u$.
One may use the change of variable $t = \frac { u } { 1 + u }$.

Using the change of variable $t = \frac { \ln x } { 2 \pi }$, demonstrate that $$\forall p \in \mathbb { N } \quad I _ { p } = \frac { e ^ { - p ^ { 2 } \pi ^ { 2 } } } { 2 \pi } \int _ { 0 } ^ { + \infty } x ^ { p - 1 } \exp \left( - \frac { \ln ^ { 2 } x } { 4 \pi ^ { 2 } } \right) \sin \left( \frac { \ln x } { 2 \pi } \right) \mathrm { d } x$$

If $a$ and $b$ are two real numbers, we denote $K _ { a , b }$ the function defined for all real $t$ by $K _ { a , b } ( t ) = \begin{cases} \frac { \mathrm { e } ^ { \mathrm { i } t b } - \mathrm { e } ^ { \mathrm { i } t a } } { 2 \mathrm { i } t } & \text { if } t \neq 0 , \\ \frac { b - a } { 2 } & \text { if } t = 0 . \end{cases}$ Show that $\int _ { - N } ^ { N } K _ { a , b } ( t ) \mathrm { d } t = \int _ { N a } ^ { N b } \operatorname { sinc } ( s ) \mathrm { d } s$.

If $n \in \mathbf{N}$, we denote by $D_n$ the improper integral $\int_0^{\pi/2} (\ln(\sin(t)))^n \mathrm{~d}t$.
Justify that, if $n \in \mathbf{N}$, the improper integral $D_n$ is convergent, then show that $$D_1 = \int_0^{\pi/2} \ln(\cos(t)) \mathrm{d}t$$

We fix $( p , q ) \in \left( \mathbf { N } ^ { * } \right) ^ { 2 }$ and set $\alpha _ { p , q } := \dfrac { p } { q }$. We define, for all $t \in \mathbf { R } _ { + }$, the application $I _ { p , q } : \mathbf { R } _ { + } \rightarrow \mathbf { R }$ by $$I _ { p , q } ( t ) := \int _ { 0 } ^ { 1 } \frac { x ^ { ( t + 1 ) \alpha _ { p , q } } } { 1 + x ^ { \alpha _ { p , q } } } d x$$ Show that, for all $( p , q ) \in \left( \mathrm { N } ^ { * } \right) ^ { 2 }$, $$S _ { p , q } = \int _ { 0 } ^ { 1 } \frac { t ^ { q - 1 } } { 1 + t ^ { p } } d t$$

Show that $$\forall u \in \mathbb{R}, \quad \forall a \in \mathbb{R}_+^*, \quad \mathrm{e}^{\frac{au^2}{2}} = \int_{-\infty}^{+\infty} \mathrm{e}^{ut - \frac{t^2}{2a}} \frac{\mathrm{~d}t}{\sqrt{2\pi a}}$$

Let $\alpha \in (0,1)$ and $\beta = \log_e(1-\alpha)$. Let $P_n(x) = x + \frac{x^2}{2} + \frac{x^3}{3} + \ldots + \frac{x^n}{n}$, $x \in (0,1)$. Then the integral $\int_0^{\alpha} \frac{t^{50}}{1-t}\,dt$ is equal to
(1) $\beta - P_{50}(\alpha)$
(2) $-\beta + P_{50}(\alpha)$
(3) $P_{50}(\alpha) - \beta$
(4) $\beta + P_{50}(\alpha)$