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1) calculate ∫_(−∞) ^∞ (dx/(1+ix)) and ∫_(−∞) ^∞ (dx/(1−ix)) 2)deduce the value of ∫_(−∞) ^∞ (dx/(1+x^2 )) 3)calculate ∫_(−∞) ^∞ (dx/(1+ix^2 )) and ∫_(−∞) ^∞ (dx/(1−ix^2 )) 4)deduce the value of ∫_(−∞) ^∞ (dx/(1+x^4 ))

$1) c a l c u l a t e \int_{- \infty}^{\infty} \frac{d x}{1 + i x} a n d \int_{- \infty}^{\infty} \frac{d x}{1 - i x}$ $2) d e d u c e t h e v a l u e o f \int_{- \infty}^{\infty} \frac{d x}{1 + x^{2}}$ $3) c a l c u l a t e \int_{- \infty}^{\infty} \frac{d x}{1 + {i x}^{2}} a n d \int_{- \infty}^{\infty} \frac{d x}{1 - {i x}^{2}}$ $4) d e d u c e t h e v a l u e o f \int_{- \infty}^{\infty} \frac{d x}{1 + x^{4}}$

Question Number 65834 Answers: 0 Comments: 1

∀ x, y >0 B(x,y)=∫_0 ^1 t^(x−1) (1−t)^(y−1) dt Γ(x)=∫_0 ^∞ t^(x−1) e^(−t) dt 1) show that ∀ x>0 Γ(x+1)=xΓ(x) and lim_(n−>∞) ((x(x+1)......(x+n))/(n^x n!))=(1/(Γ(x))) and deduce that lim_(n−>∞) ((Γ(x+n))/(n^x Γ(n)))=1 b) Prove that if a function f satisfies f(x+1)=xf(x) et lim_(n−>∞) ((f(x+n))/(n^x f(n)))=1 then ∀ x>0 f(x)= f(1)Γ(x) 3) Show that B(x+1, y)=(x/(x+y))B(x,y) B(1,x)=(1/x) 2) Now let consider ∀ y f(x)=((B(x,y)Γ(x+y))/(Γ(y))) Show that f verify the same property as Γ ( just the both proved up ) 3) Deduce that ∀ x,y>0 B(x,y)=((Γ(x)Γ(y))/(Γ(x+y)))

$\forall x, y > 0 B (x, y) = \int_{0}^{1} t^{x - 1} {(1 - t)}^{y - 1} d t Γ (x) = \int_{0}^{\infty} t^{x - 1} e^{- t} d t$ $1) s h o w t h a t \forall x > 0 Γ (x + 1) = x Γ (x) a n d {l i m}_{n - > \infty} \frac{x (x + 1) . . . . . . (x + n)}{n^{x} n!} = \frac{1}{Γ (x)}$ $a n d d e d u c e t h a t {l i m}_{n - > \infty} \frac{Γ (x + n)}{n^{x} Γ (n)} = 1$ $b) P r o v e t h a t i f a f u n c t i o n f s a t i s f i e s f (x + 1) = x f (x) e t {l i m}_{n - > \infty} \frac{f (x + n)}{n^{x} f (n)} = 1 t h e n \forall x > 0 f (x) = f (1) Γ (x)$ $3) S h o w t h a t B (x + 1, y) = \frac{x}{x + y} B (x, y) B (1, x) = \frac{1}{x}$ $2) N o w l e t c o n s i d e r \forall y f (x) = \frac{B (x, y) Γ (x + y)}{Γ (y)}$ $S h o w t h a t f v e r i f y t h e s a m e p r o p e r t y a s Γ (j u s t t h e b o t h p r o v e d u p)$ $3) D e d u c e t h a t \forall x, y > 0 B (x, y) = \frac{Γ (x) Γ (y)}{Γ (x + y)}$

Question Number 65828 Answers: 0 Comments: 1

Let go toward a rational order of derivation Part 1 : What′s that special factor Let n , p and k three integer different of zero We state J_(n,k) (p)=∫_0 ^1 (1−x^n )^(p+(k/n)) dx and C_n (p)=Π_(k=0) ^(n−1) J_(n,k) (p) 1) a) Calculate C_1 (p) b) Prove that J_(n,k) (p)=(1/n)B((1/n),p+1+(k/n) ) and explicit C_n (p)in terms of n and p 2) Deduce that ∀ n>0 there exist a real a_n such as (na_n )^n C_n (p)= (1/(p+1)) 3) Study the convergence of the result suite (a_n )_n .Then show that lim_(n−>∞) na_n =1 Part 2: the rational order of derivation Let f ∈ C^1 (R,R) . We consider I_(1/n) (f) a function defined on R_+ by I_(1/n) (f)(x)= a_n ∫_0 ^x ((f(t))/((x−t)^(1−(1/n)) ))dt and D_(1/n) (f) = (I_(1/n) (f))^((1)) 1) a _ Prove that I_((1/n) ) (f)(x)= na_n x^(1/n) ∫_0 ^1 f(x(1−v^n ))dv then find D_(1/2) (t) b) Show that ∀ f∈C^1 (R,R) ∀ x∈R_(+ ) D_(1/n) (f)(x)= I_(1/n) (f)(x) + ((f(0))/((πx)^(1−(1/n)) )) 2)∀ p integer and k∈{0,...,n−1} explicit I_(1/n) (t^(p+(k/n)) ) in term of I_(n,k) (p) b) Prove that for polynomial function f the n− th composition I_(1/n) ._ ....I_(1/n) (f)(x)=∫_0 ^x f(t)dt , c) Deduce that ∀ f polynomial the function g =f −f(0) verify D_(1/n) ......D_(1/n) (g)(x) = g(x) 3) Widen that two formulas to all function that can be developp into integer serie 4) Try to find the relation between D_(1/n) .I_(1/n) (f) , I_(1/n) .D_(1/n) (f), and f 4) Show that ∀ x∈R_+ lim_(n−>∞) I_(1/n) (f)(x)= ∫_0 ^x f(t)dt pour g=f−f(0) lim_(n−>∞) D_(1/n) (g)(x)= g(x) conclusion the derivative of the function I_α (f) defined on R_+ by I_α (f)(x)= a_n ∫_0 ^x f(t)(x−t)^((1/n)−1) dt is called the derivative of order α

$L e t g o t o w a r d a r a t i o n a l o r d e r o f d e r i v a t i o n$ $P a r t 1 : {W h a t}^{'} s t h a t s p e c i a l f a c t o r$ $L e t n, p a n d k t h r e e i n t e g e r d i f f e r e n t o f z e r o$ $W e s t a t e J_{n, k} (p) = \int_{0}^{1} {(1 - x^{n})}^{p + \frac{k}{n}} d x a n d C_{n} (p) = \underset{k = 0}{\prod^{n - 1}} J_{n, k} (p)$ $1) a) C a l c u l a t e C_{1} (p)$ $b) P r o v e t h a t J_{n, k} (p) = \frac{1}{n} B (\frac{1}{n}, p + 1 + \frac{k}{n}) a n d e x p l i c i t C_{n} (p) i n t e r m s o f n a n d p$ $2) D e d u c e t h a t \forall n > 0 t h e r e e x i s t a r e a l a_{n} s u c h a s {({n a}_{n})}^{n} C_{n} (p) = \frac{1}{p + 1}$ $3) S t u d y t h e c o n v e r g e n c e o f t h e r e s u l t s u i t e {(a_{n})}_{n} . T h e n s h o w t h a t {l i m}_{n - > \infty} {n a}_{n} = 1$ $P a r t 2 : t h e r a t i o n a l o r d e r o f d e r i v a t i o n$ $L e t f \in C^{1} (R, R) . W e c o n s i d e r I_{\frac{1}{n}} (f) a f u n c t i o n d e f i n e d o n R_{+} b y$ $I_{\frac{1}{n}} (f) (x) = a_{n} \int_{0}^{x} \frac{f (t)}{{(x - t)}^{1 - \frac{1}{n}}} d t a n d D_{\frac{1}{n}} (f) = {(I_{\frac{1}{n}} (f))}^{(1)}$ $1) a_P r o v e t h a t I_{\frac{1}{n}} (f) (x) = {n a}_{n} x^{\frac{1}{n}} \int_{0}^{1} f (x (1 - v^{n})) d v t h e n f i n d D_{\frac{1}{2}} (t)$ $b) S h o w t h a t \forall f \in C^{1} (R, R) \forall x \in R_{+} D_{\frac{1}{n}} (f) (x) = I_{\frac{1}{n}} (f) (x) + \frac{f (0)}{{(π x)}^{1 - \frac{1}{n}}}$ $2) \forall p i n t e g e r a n d k \in {0, . . ., n - 1} e x p l i c i t I_{\frac{1}{n}} (t^{p + \frac{k}{n}}) i n t e r m o f I_{n, k} (p)$ $b) P r o v e t h a t f o r p o l y n o m i a l f u n c t i o n f t h e n - t h c o m p o s i t i o n I_{\frac{1}{n}} ._{} . . . . I_{\frac{1}{n}} (f) (x) = \int_{0}^{x} f (t) d t,$ $c) D e d u c e t h a t \forall f p o l y n o m i a l t h e f u n c t i o n g = f - f (0) v e r i f y$ $D_{\frac{1}{n}} . . . . . . D_{\frac{1}{n}} (g) (x) = g (x)$ $3) W i d e n t h a t t w o f o r m u l a s t o a l l f u n c t i o n t h a t c a n b e d e v e l o p p i n t o i n t e g e r s e r i e$ $4) T r y t o f i n d t h e r e l a t i o n b e t w e e n D_{\frac{1}{n}} . I_{\frac{1}{n}} (f), I_{\frac{1}{n}} . D_{\frac{1}{n}} (f), a n d f$ $4) S h o w t h a t \forall x \in R_{+} {l i m}_{n - > \infty} I_{\frac{1}{n}} (f) (x) = \int_{0}^{x} f (t) d t$ $p o u r g = f - f (0) {l i m}_{n - > \infty} D_{\frac{1}{n}} (g) (x) = g (x)$ $c o n c l u s i o n$ $t h e d e r i v a t i v e o f t h e f u n c t i o n I_{α} (f) d e f i n e d o n R_{+} b y$ $I_{α} (f) (x) = a_{n} \int_{0}^{x} f (t) {(x - t)}^{\frac{1}{n} - 1} d t i s c a l l e d t h e d e r i v a t i v e o f o r d e r α$

Question Number 65827 Answers: 0 Comments: 0

Prove that ∫_0 ^1 (∫_(1/6) ^(5/6) (dv/((1−^v (√u) )^v )))du=ln2−2ln((√3)−1)

$P r o v e t h a t$ $\int_{0}^{1} (\int_{\frac{1}{6}}^{\frac{5}{6}} \frac{d v}{{(1 -^{v} \sqrt{u})}^{v}}) d u = l n 2 - 2 l n (\sqrt{3} - 1)$

Question Number 65825 Answers: 0 Comments: 0

let consider two real numbers p and such as p^2 −q^2 =pq Prove that J= ∫_0 ^∞ (dv/(^q (√((1+^q (√(v^p )) )^p ))))= 1

$l e t c o n s i d e r t w o r e a l n u m b e r s p a n d s u c h a s p^{2} - q^{2} = p q$ $P r o v e t h a t$ $J = \int_{0}^{\infty} \frac{d v}{^{q} \sqrt{{(1 +^{q} \sqrt{v^{p}})}^{p}}} = 1$

Question Number 65805 Answers: 2 Comments: 1

Prove that I_n =∫_0 ^(π/2) (dt/(1+(tant)^n )) does not depend of the term n deduces that ∫_0 ^∞ (dx/((x^(2035) +1)(x^2 +1)))=(π/4)

$P r o v e t h a t I_{n} = \int_{0}^{\frac{π}{2}} \frac{d t}{1 + {(t a n t)}^{n}} d o e s n o t d e p e n d o f t h e t e r m n$ $d e d u c e s t h a t$ $\int_{0}^{\infty} \frac{d x}{(x^{2035} + 1) (x^{2} + 1)} = \frac{π}{4}$

Question Number 65788 Answers: 0 Comments: 0

Explicit f(a.b.c)=∫_0 ^(π/2) ((sec(x−a))/(b.cosx + c.sinx)) dx

$E x p l i c i t f (a . b . c) = \int_{0}^{\frac{π}{2}} \frac{s e c (x - a)}{b . c o s x + c . s i n x} d x$

Question Number 65786 Answers: 0 Comments: 0

Shows that ∣Γ(1+ix)∣^2 =(π/(xsinh(πx))) with Γ(z)=∫_0_ ^∞ t^(z−1) e^(−t) dt Then Prove that ∫_0 ^∞ ∣Γ(1+ix)∣^2 dx =(π/4)

$S h o w s t h a t ∣ Γ (1 + i x) ∣^{2} = \frac{π}{x s i n h (π x)} w i t h Γ (z) = \int_{0_{}}^{\infty} t^{z - 1} e^{- t} d t$ $T h e n P r o v e t h a t \int_{0}^{\infty} ∣ Γ (1 + i x) ∣^{2} d x = \frac{π}{4}$

Question Number 65776 Answers: 0 Comments: 1

find ∫_(−(π/4)) ^(π/4) ((cosx)/(2+5sinx))dx

$f i n d \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{c o s x}{2 + 5 s i n x} d x$

Question Number 65775 Answers: 0 Comments: 1

calculate ∫_0 ^(2π) ((tanx)/(2+3cosx))dx

$c a l c u l a t e \int_{0}^{2 π} \frac{t a n x}{2 + 3 c o s x} d x$

Question Number 65774 Answers: 0 Comments: 0

find A_n = ∫_0 ^(2π) ((sin^2 x)/(sin^2 (((nx)/2))))dx (n>0)

$f i n d A_{n} = \int_{0}^{2 π} \frac{{s i n}^{2} x}{{s i n}^{2} (\frac{n x}{2})} d x (n > 0)$

Question Number 65773 Answers: 0 Comments: 2

let f(x) =∫_0 ^1 (dt/(1+x(√(1+t^2 )))) with x>0 1)detemine a explicit form of f(x) 2)find also g(x) =∫_0 ^1 ((√(1+t^2 ))/((1+x(√(1+t^2 )))^2 ))dt 3) find the value of integrals ∫_0 ^1 (dt/(1+2(√(1+t^2 )))) and ∫_0 ^1 (dt/((1+2(√(1+t^2 )))^2 ))

$l e t f (x) = \int_{0}^{1} \frac{d t}{1 + x \sqrt{1 + t^{2}}} w i t h x > 0$ $1) d e t e m i n e a e x p l i c i t f o r m o f f (x)$ $2) f i n d a l s o g (x) = \int_{0}^{1} \frac{\sqrt{1 + t^{2}}}{{(1 + x \sqrt{1 + t^{2}})}^{2}} d t$ $3) f i n d t h e v a l u e o f i n t e g r a l s \int_{0}^{1} \frac{d t}{1 + 2 \sqrt{1 + t^{2}}} a n d$ $\int_{0}^{1} \frac{d t}{{(1 + 2 \sqrt{1 + t^{2}})}^{2}}$

Question Number 65771 Answers: 0 Comments: 0

let X_n =∫_0 ^(π/4) sin^n xdx 1) calculate X_0 ,X_1 ,X_2 ,X_3 2) find X_n interms of n 3)find the value of ∫_0 ^(π/4) sin^8 xdx

$l e t X_{n} = \int_{0}^{\frac{π}{4}} {s i n}^{n} x d x$ $1) c a l c u l a t e X_{0}, X_{1}, X_{2}, X_{3}$ $2) f i n d X_{n} i n t e r m s o f n$ $3) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{4}} {s i n}^{8} x d x$

Question Number 65770 Answers: 0 Comments: 2

let A_n =∫_0 ^(π/2) cos^n xdx 1) calculate A_0 ,A_2 and A_3 2)calculate A_n interms of n 3) find ∫_0 ^(π/2) cos^8 xdx

$l e t A_{n} = \int_{0}^{\frac{π}{2}} {c o s}^{n} x d x$ $1) c a l c u l a t e A_{0}, A_{2} a n d A_{3}$ $2) c a l c u l a t e A_{n} i n t e r m s o f n$ $3) f i n d \int_{0}^{\frac{π}{2}} {c o s}^{8} x d x$

Question Number 65769 Answers: 0 Comments: 3

find the value of ∫_0 ^∞ (dx/((x^2 −2xcosθ +1)^2 ))

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{{(x^{2} - 2 x c o s θ + 1)}^{2}}$

Question Number 65768 Answers: 0 Comments: 1

find the value of ∫_0 ^∞ (dx/(x^2 −2(cosθ)x +1))

$f i n d t h e v a l u e o f \int_{0}^{\infty} \frac{d x}{x^{2} - 2 (c o s θ) x + 1}$

Pg 213 Pg 214 Pg 215 Pg 216 Pg 217 Pg 218 Pg 219 Pg 220 Pg 221 Pg 222

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