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....... Advanced ::::::::::★★★:::::::::: Calculus....... find the value of the infinite series:: Θ := Σ_(n=1) ^∞ (((−1)^(n−1) H_( 2n) )/(2n−1)) = ??? .......M.N.july.1970........

$. . . . . . . A d v a n c e d :::::::::: ★ ★ ★ :::::::::: C a l c u l u s . . . . . . .$ $f i n d t h e v a l u e o f t h e i n f i n i t e s e r i e s ::$ $Θ := \underset{n = 1}{\sum^{\infty}} \frac{{(- 1)}^{n - 1} H_{2 n}}{2 n - 1} = ? ? ?$ $. . . . . . . M . N . j u l y . 1970 . . . . . . . .$

Question Number 140823 Answers: 1 Comments: 1

Determine whether the improper integral converges or diverges ∫_1 ^( ∞) ((2x+7)/(7x^3 +5x^2 +1)) dx

$Determine whether the improper$ $integral converges or diverges$ $\int_{1}^{\infty} \frac{2 x + 7}{{7 x}^{3} + {5 x}^{2} + 1} dx$

Question Number 140793 Answers: 0 Comments: 0

∫(√((x+2)/e^x ))dx=...?

$\int \sqrt{\frac{x + 2}{e^{x}}} d x = . . . ?$

Question Number 140789 Answers: 2 Comments: 0

.......nice.....math.... calculate:: Θ:= Σ_(n=1) ^∞ (1/) =??

$. . . . . . . n i c e . . . . . m a t h . . . .$ $c a l c u l a t e :: Θ := \underset{n = 1}{\sum^{\infty}} \frac{1}{} = ? ?$

Question Number 140768 Answers: 3 Comments: 0

∫_(−∞) ^∞ ((x^2 +4)/(x^4 +16)) dx =?

$\int_{- \infty}^{\infty} \frac{x^{2} + 4}{x^{4} + 16} dx = ?$

Question Number 140715 Answers: 1 Comments: 0

......advanced calculus...... prove that: 𝛗:=∫_0 ^( ∞) (x^2 /(cosh^2 (x^2 )))dx=^? (((√2) −2)/4) (√π) ζ ( (1/2) ) ..............

$. . . . . . a d v a n c e d c a l c u l u s . . . . . .$ $p r o v e t h a t :$ $ϕ := \int_{0}^{\infty} \frac{x^{2}}{{c o s h}^{2} (x^{2})} d x \overset{?}{=} \frac{\sqrt{2} - 2}{4} \sqrt{π} ζ (\frac{1}{2})$ $. . . . . . . . . . . . . .$

Question Number 140703 Answers: 2 Comments: 0

Question Number 140684 Answers: 2 Comments: 0

∫_0 ^π cos^n (x)∙cos (nx)dx=(π/2^n )

$\int_{0}^{π} \cos^{n} (x) \cdot \cos (n x) d x = \frac{π}{2^{n}}$

Question Number 140685 Answers: 0 Comments: 0

Question Number 140687 Answers: 0 Comments: 2

Question Number 140645 Answers: 0 Comments: 0

Question Number 140639 Answers: 1 Comments: 0

let U_n =∫_0 ^∞ ((x^n logx)/((x^2 +1)^2 ))dx 1) explicite U_n 2) fond nature of Σ U_n (n integr natural)

$l e t U_{n} = \int_{0}^{\infty} \frac{x^{n} l o g x}{{(x^{2} + 1)}^{2}} d x$ $1) e x p l i c i t e U_{n}$ $2) f o n d n a t u r e o f Σ U_{n}$ $(n i n t e g r n a t u r a l)$

Question Number 140634 Answers: 0 Comments: 0

find ∫_(−∞) ^(+∞) ((cos(2sinx))/((x^2 −x+1)^2 ))dx

$f i n d \int_{- \infty}^{+ \infty} \frac{c o s (2 s i n x)}{{(x^{2} - x + 1)}^{2}} d x$

Question Number 140635 Answers: 1 Comments: 0

find ∫_(−∞) ^(+∞) ((sin(2cosx))/((x^2 −x+1)^2 ))dx

$f i n d \int_{- \infty}^{+ \infty} \frac{s i n (2 c o s x)}{{(x^{2} - x + 1)}^{2}} d x$

Question Number 140615 Answers: 1 Comments: 0

Find the area common to the curve y^2 = 12x and x^2 +y^2 = 24x .

$Find the area common$ $to the curve y^{2} = 12 x and$ $x^{2} + y^{2} = 24 x .$

Question Number 140614 Answers: 2 Comments: 0

∫ _0^(π/2) ln (sin x) sec^2 x dx =?

$\int_{0}^{\frac{π}{2}} \ln (\sin x) \sec^{2} x dx = ?$

Question Number 140588 Answers: 1 Comments: 0

.......Advanced ....★★★....Calculus....... evaluation the value of : 𝛗 :=∫_0 ^( (π/2)) sin^2 (x).ln(sin(x))dx solution:: ξ (a):=∫_0 ^( (π/2)) sin^(2+a) (x)dx =(1/2)β (((3+a)/2) ,(1/2)) :=(1/2)(((Γ(((3+a)/2))Γ((1/2)))/(Γ(2+(a/2))))).......✓ 𝛗:= ξ ′ (0) ..............✓ :=(1/2) (√π) (( Γ′(((3+a)/2)).Γ(2+(a/2))−Γ(((3+a)/2)).Γ′(2+(a/2)))/(Γ^2 (2+(a/2)))) ∣_(a=0) :=(1/2)(√π) ((Γ′((3/2))−Γ((3/2)).Γ′(2))/(( Γ^2 (2):=1 ))) :=(1/2)(√π) ((ψ((3/2))Γ((3/2))−Γ((3/2)).ψ(2))/1) := ((√π)/4){ (2−γ−2ln(2)−(1−γ)} :=((√π)/4)(1−ln(4))=(√π) ln(((e/4))^(1/4) )

$. . . . . . . A d v a n c e d . . . . ★ ★ ★ . . . . C a l c u l u s . . . . . . .$ $e v a l u a t i o n t h e v a l u e o f :$ $ϕ := \int_{0}^{\frac{π}{2}} {s i n}^{2} (x) . l n (s i n (x)) d x$ $s o l u t i o n ::$ $ξ (a) := \int_{0}^{\frac{π}{2}} {s i n}^{2 + a} (x) d x = \frac{1}{2} β (\frac{3 + a}{2}, \frac{1}{2})$ $:= \frac{1}{2} (\frac{Γ (\frac{3 + a}{2}) Γ (\frac{1}{2})}{Γ (2 + \frac{a}{2})}) . . . . . . . ✓$ $ϕ := ξ^{'} (0) . . . . . . . . . . . . . . ✓$ $:= \frac{1}{2} \sqrt{π} \frac{Γ^{'} (\frac{3 + a}{2}) . Γ (2 + \frac{a}{2}) - Γ (\frac{3 + a}{2}) . Γ^{'} (2 + \frac{a}{2})}{Γ^{2} (2 + \frac{a}{2})} ∣_{a = 0}$ $:= \frac{1}{2} \sqrt{π} \frac{Γ^{'} (\frac{3}{2}) - Γ (\frac{3}{2}) . Γ^{'} (2)}{(Γ^{2} (2) := 1)}$ $:= \frac{1}{2} \sqrt{π} \frac{ψ (\frac{3}{2}) Γ (\frac{3}{2}) - Γ (\frac{3}{2}) . ψ (2)}{1}$ $:= \frac{\sqrt{π}}{4} {(2 - γ - 2 l n (2) - (1 - γ)}$ $:= \frac{\sqrt{π}}{4} (1 - l n (4)) = \sqrt{π} l n (\sqrt[4]{\frac{e}{4}})$

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