calculate-1-2x-x-x-4-dx-

Question Number 61041 by mathsolverby Abdo last updated on 28/May/19

c a l c u l a t e \int_{1}^{+ \infty} \frac{[2 x] - [x]}{x^{4}} d x

Commented by maxmathsup by imad last updated on 29/May/19

let A =∫_1 ^(+∞) (([2x]−[x])/x^4 ) dx ⇒ A =∫_1 ^(+∞) (([2x])/x^4 ) dx −∫_1 ^(+∞) (([x])/x^4 ) dx ∫_1 ^(+∞) (([2x])/x^4 ) dx =_(2x=t) 2^4 ∫_2 ^(+∞) (([t])/t^4 ) (dt/2) =8 ∫_2 ^(+∞) (([t])/t^4 ) dt =8 ( ∫_1 ^(+∞) (([t])/t^4 ) dt −∫_1 ^2 (([t])/t^4 ) dt) =8 ∫_1 ^(+∞) (([t])/t^4 ) dt −8 ∫_1 ^2 (dt/t^4 ) ∫_1 ^2 (dt/t^4 ) =[(1/(−4+1)) t^(−4+1) ]_1 ^2 =[−(1/3) t^(−3) ]_1 ^2 =−(1/3)(2^(−3) −1)=(1/3)(1−(1/8))=(1/3) (7/8) ⇒ ∫_1 ^(+∞) (([2x])/x^4 ) dx =8 ∫_1 ^(+∞) (([t])/t^4 )dt−(7/3) ⇒A =7 ∫_1 ^(+∞) (([x])/x^4 )dx −(7/3) ∫_1 ^(+∞) (([x])/x^4 ) dx =lim_(n→+∞) Σ_(k=1) ^(n−1) ∫_k ^(k+1) (([x])/x^4 ) dx =Σ_(n=1) ^∞ ∫_n ^(n+1) (([x])/x^4 ) dx =Σ_(n=1) ^∞ n ∫_n ^(n+1) (dx/x^4 ) =Σ_(n=1) ^∞ n[(1/(−4+1)) x^(−4+1) ]_n ^(n+1) =Σ_(n=1) ^∞ n[−(1/3) (1/x^3 )]_n ^(n+1) =−(1/3) Σ_(n=1) ^∞ n((1/((n+1)^3 )) −(1/n^3 )) =(1/3) Σ_(n=1) ^∞ (1/n^2 ) −(1/3) Σ_(n=1) ^∞ (n/((n+1)^3 )) =(1/3)ξ(2)−(1/3)Σ_(n=1) ^∞ ((n+1−1)/((n+1)^3 )) =(1/3)ξ(2)−(1/3) Σ_(n=1) ^∞ (1/((n+1)^2 )) +(1/3)Σ_(n=1) ^∞ (1/((n+1)^3 )) Σ_(n=1) ^∞ (1/((n+1)^2 ))=Σ_(n=2) ^∞ (1/n^2 ) =ξ(2)−1 Σ_(n=1) ^∞ (1/((n+1)^3 )) =Σ_(n=2) ^∞ (1/n^3 ) =ξ(3)−1 ⇒ ∫_1 ^(+∞) (([x])/x^4 ) dx = (1/3)ξ(2) −(1/3)(ξ(2)−1) +(1/3) (ξ(3)−1) =(1/3) +(1/3)ξ(3)−(1/3) =(1/3)ξ(3) ⇒ A =(7/3)ξ(3)−(7/3) .

l e t A = \int_{1}^{+ \infty} \frac{[2 x] - [x]}{x^{4}} d x \Rightarrow A = \int_{1}^{+ \infty} \frac{[2 x]}{x^{4}} d x - \int_{1}^{+ \infty} \frac{[x]}{x^{4}} d x

\int_{1}^{+ \infty} \frac{[2 x]}{x^{4}} d x =_{2 x = t} 2^{4} \int_{2}^{+ \infty} \frac{[t]}{t^{4}} \frac{d t}{2} = 8 \int_{2}^{+ \infty} \frac{[t]}{t^{4}} d t

= 8 (\int_{1}^{+ \infty} \frac{[t]}{t^{4}} d t - \int_{1}^{2} \frac{[t]}{t^{4}} d t) = 8 \int_{1}^{+ \infty} \frac{[t]}{t^{4}} d t - 8 \int_{1}^{2} \frac{d t}{t^{4}}

\int_{1}^{2} \frac{d t}{t^{4}} = {[\frac{1}{- 4 + 1} t^{- 4 + 1}]}_{1}^{2} = {[- \frac{1}{3} t^{- 3}]}_{1}^{2} = - \frac{1}{3} (2^{- 3} - 1) = \frac{1}{3} (1 - \frac{1}{8}) = \frac{1}{3} \frac{7}{8} \Rightarrow

\int_{1}^{+ \infty} \frac{[2 x]}{x^{4}} d x = 8 \int_{1}^{+ \infty} \frac{[t]}{t^{4}} d t - \frac{7}{3} \Rightarrow A = 7 \int_{1}^{+ \infty} \frac{[x]}{x^{4}} d x - \frac{7}{3}

\int_{1}^{+ \infty} \frac{[x]}{x^{4}} d x = {l i m}_{n \to + \infty} \sum_{k = 1}^{n - 1} \int_{k}^{k + 1} \frac{[x]}{x^{4}} d x

= \sum_{n = 1}^{\infty} \int_{n}^{n + 1} \frac{[x]}{x^{4}} d x = \sum_{n = 1}^{\infty} n \int_{n}^{n + 1} \frac{d x}{x^{4}} = \sum_{n = 1}^{\infty} n {[\frac{1}{- 4 + 1} x^{- 4 + 1}]}_{n}^{n + 1}

= \sum_{n = 1}^{\infty} n {[- \frac{1}{3} \frac{1}{x^{3}}]}_{n}^{n + 1} = - \frac{1}{3} \sum_{n = 1}^{\infty} n (\frac{1}{{(n + 1)}^{3}} - \frac{1}{n^{3}})

= \frac{1}{3} \sum_{n = 1}^{\infty} \frac{1}{n^{2}} - \frac{1}{3} \sum_{n = 1}^{\infty} \frac{n}{{(n + 1)}^{3}}

= \frac{1}{3} ξ (2) - \frac{1}{3} \sum_{n = 1}^{\infty} \frac{n + 1 - 1}{{(n + 1)}^{3}} = \frac{1}{3} ξ (2) - \frac{1}{3} \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}} + \frac{1}{3} \sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{3}}

\sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{2}} = \sum_{n = 2}^{\infty} \frac{1}{n^{2}} = ξ (2) - 1

\sum_{n = 1}^{\infty} \frac{1}{{(n + 1)}^{3}} = \sum_{n = 2}^{\infty} \frac{1}{n^{3}} = ξ (3) - 1 \Rightarrow

\int_{1}^{+ \infty} \frac{[x]}{x^{4}} d x = \frac{1}{3} ξ (2) - \frac{1}{3} (ξ (2) - 1) + \frac{1}{3} (ξ (3) - 1)

= \frac{1}{3} + \frac{1}{3} ξ (3) - \frac{1}{3} = \frac{1}{3} ξ (3) \Rightarrow A = \frac{7}{3} ξ (3) - \frac{7}{3} .

Answered by perlman last updated on 28/May/19

∫_1 ^(+∞) (([2x]−[x])/x^4 )dx=Σ_(k=1) ^(+∞) ∫_((k+1)/2) ^((k+2)/2) (([2x])/x^(4 ) )dx−Σ_(k=1) ^(+∞) ∫_k ^(k+1) (([x])/x^4 )dx [2x]=k+1 ∀x∈[((k+1)/2),((k+2)/2)[ [x]=k ∀x∈[k,k+1[ ∫_1 ^(+∞) (([2x]−[x])/x^4 )=lim_(n→∞) Σ_(k=1) ^n [∫_((k+1)/2) ^((k+2)/2) ((k+1)/x^4 )dx−∫_k ^(k+1) (k/x^(4 ) )dx] =lim_(n→∞) Σ_(k=1) ^(k=n) [((k+1)/(−3(((k+2)/2))^3 ))+((k+1)/(3(((k+1)/2))^3 ))+(k/(3(k+1)^3 ))−(k/(3k^3 ))] =lim_(n→∞) Σ_(k=1) ^(k=n) [((−8(k+1))/( 3(k+2)^3 ))+(8/(3(k+1)^2 ))+(k/(3(k+1)^3 ))−(1/(3k^2 ))] we use (x/((x+1)^3 ))=(1/((x+1)^2 ))−(1/((x+1)^3 )) andlim_(x→∞) Σ_(k=1) ^x (1/k^z )=ζ(z) we find limΣ_(k=1) ^n [((−8)/(3(k+2)^2 ))+(8/(3(k+2)^3 ))+(8/(3(k+1)^2 ))+(1/(3(k+1)^2 ))−(1/(3(k+1)^3 ))−(1/(3k^2 ))] =lim [(8/(12))−(8/(3(n+2)^2 ))+(8/3)Σ_(k=1) ^n (1/((k+2)^3 ))+(1/(3(n+1)^2 ))−(1/3)−(1/3)Σ_(k=1) ^n (1/((k+1)^3 ))] =(8/(12))−(1/3)+(8/3)Σ_(k=1) ^∞ (1/((k+2)^3 ))−(1/3)Σ_(k=1) ^∞ (1/((k+1)^3 ))=(1/3)+(8/3)(ζ(3)−1−(1/8))−(1/3)(ζ(3)−1) =(1/3)+(7/3)ζ(3)−3+(1/3)=(7/3)ζ(3)−(7/3)=(7/3)(ζ(3)−1)

\int_{1}^{+ \infty} \frac{[2 x] - [x]}{x^{4}} d x = \underset{k = 1}{\sum^{+ \infty}} \int_{\frac{k + 1}{2}}^{\frac{k + 2}{2}} \frac{[2 x]}{x^{4}} d x - \underset{k = 1}{\sum^{+ \infty}} \underset{k}{\int^{k + 1}} \frac{[x]}{x^{4}} d x

[2 x] = k + 1 \forall x \in [\frac{k + 1}{2}, \frac{k + 2}{2} [

[x] = k \forall x \in [k, k + 1 [

\underset{1}{\int^{+ \infty}} \frac{[2 x] - [x]}{x^{4}} = \lim_{n \to \infty} \underset{k = 1}{\sum^{n}} [\underset{\frac{k + 1}{2}}{\int^{\frac{k + 2}{2}}} \frac{k + 1}{x^{4}} d x - \underset{k}{\int^{k + 1}} \frac{k}{x^{4}} d x]

= \lim_{n \to \infty} \underset{k = 1}{\sum^{k = n}} [\frac{k + 1}{- 3 {(\frac{k + 2}{2})}^{3}} + \frac{k + 1}{3 {(\frac{k + 1}{2})}^{3}} + \frac{k}{3 {(k + 1)}^{3}} - \frac{k}{3 k^{3}}]

= \lim_{n \to \infty} \underset{k = 1}{\sum^{k = n}} [\frac{- 8 (k + 1)}{3 {(k + 2)}^{3}} + \frac{8}{3 {(k + 1)}^{2}} + \frac{k}{3 {(k + 1)}^{3}} - \frac{1}{3 k^{2}}]

w e u s e \frac{x}{{(x + 1)}^{3}} = \frac{1}{{(x + 1)}^{2}} - \frac{1}{{(x + 1)}^{3}} a n d \lim_{x \to \infty} \underset{k = 1}{\sum^{x}} \frac{1}{k^{z}} = ζ (z)

w e f i n d

l i m \underset{k = 1}{\sum^{n}} [\frac{- 8}{3 {(k + 2)}^{2}} + \frac{8}{3 {(k + 2)}^{3}} + \frac{8}{3 {(k + 1)}^{2}} + \frac{1}{3 {(k + 1)}^{2}} - \frac{1}{3 {(k + 1)}^{3}} - \frac{1}{3 k^{2}}]

= l i m [\frac{8}{12} - \frac{8}{3 {(n + 2)}^{2}} + \frac{8}{3} \underset{k = 1}{\sum^{n}} \frac{1}{{(k + 2)}^{3}} + \frac{1}{3 {(n + 1)}^{2}} - \frac{1}{3} - \frac{1}{3} \underset{k = 1}{\sum^{n}} \frac{1}{{(k + 1)}^{3}}]

= \frac{8}{12} - \frac{1}{3} + \frac{8}{3} \underset{k = 1}{\sum^{\infty}} \frac{1}{{(k + 2)}^{3}} - \frac{1}{3} \underset{k = 1}{\sum^{\infty}} \frac{1}{{(k + 1)}^{3}} = \frac{1}{3} + \frac{8}{3} (ζ (3) - 1 - \frac{1}{8}) - \frac{1}{3} (ζ (3) - 1)

= \frac{1}{3} + \frac{7}{3} ζ (3) - 3 + \frac{1}{3} = \frac{7}{3} ζ (3) - \frac{7}{3} = \frac{7}{3} (ζ (3) - 1)

Commented by Tawa1 last updated on 28/May/19

Commented by Tawa1 last updated on 28/May/19

Please sir, sum the series .

Commented by perlman last updated on 29/May/19

=Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2))) (1/(x(x+1)(x+2)))=(1/(2x))−(1/(x+1))+(1/(2(x+2))) Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2)))=lim_(x→+∞) Σ_(k=1) ^x [(1/(2k))−(1/(k+1))+(1/(2(k+2)))] =lim_(x→+∞) Σ_(k=1) ^x (1/(2k))−Σ_(k=1) ^x (1/(k+1))+Σ(1/(2(k+2))) put r=k+1 in the seconde sum and v=k+2 in the 3rd =lim_(x→+∞) Σ_(k=1) ^x (1/(2k))−Σ_(r=2) ^(x+1) (1/r)+Σ_(v=3) ^(x+2) (1/(2v))=lim_(x→+∞) ((1/2)+(1/4)+Σ_(k=3) ^x (1/(2k))−Σ_2 ^(x+1) (1/k)+Σ_3 ^x (1/(2k))+(1/(2x+2))+(1/(2(x+2)))) =lim_(x→+∞) ((3/4)+2Σ_(k=3) ^x (1/(2k))−(1/2)−Σ_(k=3) ^x (1/k)−(1/(x+1))+(1/(2x+2))+(1/(2x+4))) =lim_(x→+∞) ((1/4)+Σ_3 ^x (1/k)−Σ_3 ^x (1/k)−(1/(2x+2))+(1/(2x+4))) =lim_(x→+∞) ((1/4)−(1/(2x+2))+(1/(2x+4)))=(1/4)

= \underset{k = 1}{\sum^{+ \infty}} \frac{1}{k (k + 1) (k + 2)}

\frac{1}{x (x + 1) (x + 2)} = \frac{1}{2 x} - \frac{1}{x + 1} + \frac{1}{2 (x + 2)}

\underset{k = 1}{\sum^{+ \infty}} \frac{1}{k (k + 1) (k + 2)} = \lim_{x \to + \infty} \underset{k = 1}{\sum^{x}} [\frac{1}{2 k} - \frac{1}{k + 1} + \frac{1}{2 (k + 2)}]

= \lim_{x \to + \infty} \underset{k = 1}{\sum^{x}} \frac{1}{2 k} - \underset{k = 1}{\sum^{x}} \frac{1}{k + 1} + Σ \frac{1}{2 (k + 2)}

p u t r = k + 1 i n t h e s e c o n d e s u m

a n d v = k + 2 i n t h e 3 r d

= \lim_{x \to + \infty} \underset{k = 1}{\sum^{x}} \frac{1}{2 k} - \underset{r = 2}{\sum^{x + 1}} \frac{1}{r} + \underset{v = 3}{\sum^{x + 2}} \frac{1}{2 v} = \lim_{x \to + \infty} (\frac{1}{2} + \frac{1}{4} + \underset{k = 3}{\sum^{x}} \frac{1}{2 k} - \underset{2}{\sum^{x + 1}} \frac{1}{k} + \underset{3}{\sum^{x}} \frac{1}{2 k} + \frac{1}{2 x + 2} + \frac{1}{2 (x + 2)})

= \lim_{x \to + \infty} (\frac{3}{4} + 2 \underset{k = 3}{\sum^{x}} \frac{1}{2 k} - \frac{1}{2} - \underset{k = 3}{\sum^{x}} \frac{1}{k} - \frac{1}{x + 1} + \frac{1}{2 x + 2} + \frac{1}{2 x + 4})

= \lim_{x \to + \infty} (\frac{1}{4} + \underset{3}{\sum^{x}} \frac{1}{k} - \underset{3}{\sum^{x}} \frac{1}{k} - \frac{1}{2 x + 2} + \frac{1}{2 x + 4})

= \lim_{x \to + \infty} (\frac{1}{4} - \frac{1}{2 x + 2} + \frac{1}{2 x + 4}) = \frac{1}{4}

Commented by maxmathsup by imad last updated on 29/May/19

look sir Σ_(k=1) ^n (1/(k(k+1)(k+2))) =(1/(1.2.3)) +(1/(2.3.4)) +(1/(3.4.5)) +...so this isn t the sum given at the question...!

l o o k s i r \sum_{k = 1}^{n} \frac{1}{k (k + 1) (k + 2)} = \frac{1}{1.2.3} + \frac{1}{2.3.4} + \frac{1}{3.4.5} + \dots s o t h i s i s n t

t h e s u m g i v e n a t t h e q u e s t i o n \dots!

Commented by perlman last updated on 29/May/19

2nd un=(1/(n(n+1)(n+2))) serie of therm un convergent let f(x)=Σ_(n=1 ) ^(+∞) (x^(n+2) /(n(n+1)(n+2))) f^((1)) (x)=Σ_(n=1) ^(+∞) (x^(n+1) /(n(n+1))) f^((2)) (x)=Σ_(n=1) ^(+∞) (x^n /n)=−ln(1−x)“devlop of ln(1−z) in power series” f^((1)) (x)=∫f^((1)) (x)dx=∫−ln(1−x)dx=−∫ln(1−x)dx=(1−x)ln(1−x)+x“caus f^((1)) (0)=0” f(x)=∫f^((1)) dx=∫(1−x)ln(1−x)+x dx=((−(1−x)^2 ln(1−x))/2)+(((1−x)^2 )/4)+(x^2 /2)+c f(0)=0===≥(1/4)+c=0 so c=−(1/4) f(x)=((−(1−x)^2 ln(1−x))/2)+(((1−x)^2 )/4)+(x^2 /2)−(1/4) for x=1 we get f(1)=Σ_(k=1) ^(+∞) (1/(k(k+1)(k+2)))=(1/2)−(1/4)=(1/4)

2 n d

u n = \frac{1}{n (n + 1) (n + 2)}

s e r i e o f t h e r m u n c o n v e r g e n t

l e t f (x) = \underset{n = 1}{\sum^{+ \infty}} \frac{x^{n + 2}}{n (n + 1) (n + 2)}

f^{(1)} (x) = \underset{n = 1}{\sum^{+ \infty}} \frac{x^{n + 1}}{n (n + 1)}

f^{(2)} (x) = \underset{n = 1}{\sum^{+ \infty}} \frac{x^{n}}{n} = - l n (1 - x) “ d e v l o p o f l n (1 - z) i n p o w e r {s e r i e s}^{″}

f^{(1)} (x) = \int f^{(1)} (x) d x = \int - l n (1 - x) d x = - \int l n (1 - x) d x = (1 - x) l n (1 - x) + x “ c a u s f^{(1)} (0) = 0^{″}

f (x) = \int f^{(1)} d x = \int (1 - x) l n (1 - x) + x d x = \frac{- {(1 - x)}^{2} l n (1 - x)}{2} + \frac{{(1 - x)}^{2}}{4} + \frac{x^{2}}{2} + c

f (0) = 0 ===⩾ \frac{1}{4} + c = 0 s o c = - \frac{1}{4}

f (x) = \frac{- {(1 - x)}^{2} l n (1 - x)}{2} + \frac{{(1 - x)}^{2}}{4} + \frac{x^{2}}{2} - \frac{1}{4}

f o r x = 1 w e g e t f (1) = \underset{k = 1}{\sum^{+ \infty}} \frac{1}{k (k + 1) (k + 2)} = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}

Commented by perlman last updated on 29/May/19

y^{'} r e w e l c o m

Commented by Tawa1 last updated on 29/May/19

Wow. God bless you sir. Sir, Does the series has sum of nth term

Wow . God bless you sir .

Sir, Does the series has sum of nth term

Commented by Tawa1 last updated on 29/May/19

I want to know if the sum to nth term correspond to the given series. that is, using the sum of nth term formular to verify. S_1 , S_2 , S_3 ... if it correspond

I want to know if the sum to nth term correspond to the given series .

that is, using the sum of nth term formular to verify . S_{1}, S_{2}, S_{3} \dots

if it correspond

Commented by perlman last updated on 29/May/19

sir ther is mistak in my answer i will fix it

s i r t h e r i s m i s t a k i n m y a n s w e r i w i l l f i x i t

Commented by perlman last updated on 29/May/19

the sum is Σ_(k=1) ^(+∞) (1/(3k(3k−1)(3k−2))).and notΣ(1/(k(k+1)(k+2)))sorry for this/mistak

t h e s u m i s \underset{k = 1}{\sum^{+ \infty}} \frac{1}{3 k (3 k - 1) (3 k - 2)} . a n d n o t Σ \frac{1}{k (k + 1) (k + 2)} s o r r y f o r t h i s / m i s t a k

Commented by Tawa1 last updated on 29/May/19

Okay sir, when you check. Tell me sum of the first n terms

Okay sir, when you check . Tell me sum of the first n terms

Commented by tanmay last updated on 29/May/19

T_n =(1/([1+(n−1)3][2+(n−1)3][3+(n−1)3])) T_n =(1/((3n−2)((3n−1)(3n))) S_n =C−(1/(3×1×(3n−1)3n)) S_1 =C−(1/(18))=T_1 =(1/6) C=(1/6)+(1/(18))=((3+1)/(18))=(2/9) S_n =(2/9)−(1/(9n(3n−1))) S_∞ =(2/9) pls check...

T_{n} = \frac{1}{[1 + (n - 1) 3] [2 + (n - 1) 3] [3 + (n - 1) 3]}

T_{n} = \frac{1}{(3 n - 2) ((3 n - 1) (3 n)}

S_{n} = C - \frac{1}{3 \times 1 \times (3 n - 1) 3 n}

S_{1} = C - \frac{1}{18} = T_{1} = \frac{1}{6}

C = \frac{1}{6} + \frac{1}{18} = \frac{3 + 1}{18} = \frac{2}{9}

S_{n} = \frac{2}{9} - \frac{1}{9 n (3 n - 1)}

S_{\infty} = \frac{2}{9}

p l s c h e c k \dots

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