An interesting definite integral $int_0^1(1+x+x^2+x^3+cdotcdotcdot+x^n-1)^2 (1+4x+7x^2+cdotcdotcdot+(3n-2)x^n-1)~dx=n^3$
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How to prove
$~~ forall ninmathbbN^+$,
beginalignI_n=int_0^1(1+x+x^2+x^3+cdotcdotcdot+x^n-1)^2 (1+4x+7x^2+cdotcdotcdot+(3n-2)x^n-1)~dx=n^3.endalign
My Try:
Define $displaystyle S(n)=sum_k=0^n-1x^k=1+x+x^2+x^3+cdotcdotcdot+x^n-1=fracx^n-1x-1$. Then,
beginalignfracddxS(n)=S'(n)=1+2x+3x^2+cdotcdotcdot(n-1)x^n-2=sum_k=0^n-1kx^k-1.endalign
Therefore,
beginalignI_n=int_0^1 S^2(n)left(3S'(n+1)-2S(n)right)~dxendalign
beginalign=3int_0^1 S^2(n)S'(n+1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)(S'(n)+nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)~d(S(n))+3int_0^1 S^2(n)(nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=n^3-1+int_0^1 S^2(n)(3nx^n-1-2S(n))~dxendalign
beginalign=n^3-1+int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dxendalign
So the question becomes:
Prove beginalignI'=int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dx=1.endalign
beginalignI'=int_0^1 frac3nx^n-1(x^n-1)^2(x-1)^2-frac2(x^n-1)^3(x-1)^3~dxendalign
beginalign=int_0^1 frac(x-1)^2left(frac d dx (x^n-1)^3right)-2(x^n-1)^3(x-1)(x-1)^4~dxendalign
beginalign=int_0^1 frac d dx left(frac(x^n-1)^3(x-1)^2right)~dxendalign
beginalign=lim_x to 1 frac(x^n-1)^3(x-1)^2-frac(0^n-1)^3(0-1)^2endalign
beginaligntherefore I'=1.endalign
beginaligntherefore I_n=n^3.endalign
There MUST be other BETTER ways evaluating $I_n$.
Could anyone give me some better solutions? Thanks.
calculus integration definite-integrals
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
add a comment |Â
up vote
12
down vote
favorite
How to prove
$~~ forall ninmathbbN^+$,
beginalignI_n=int_0^1(1+x+x^2+x^3+cdotcdotcdot+x^n-1)^2 (1+4x+7x^2+cdotcdotcdot+(3n-2)x^n-1)~dx=n^3.endalign
My Try:
Define $displaystyle S(n)=sum_k=0^n-1x^k=1+x+x^2+x^3+cdotcdotcdot+x^n-1=fracx^n-1x-1$. Then,
beginalignfracddxS(n)=S'(n)=1+2x+3x^2+cdotcdotcdot(n-1)x^n-2=sum_k=0^n-1kx^k-1.endalign
Therefore,
beginalignI_n=int_0^1 S^2(n)left(3S'(n+1)-2S(n)right)~dxendalign
beginalign=3int_0^1 S^2(n)S'(n+1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)(S'(n)+nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)~d(S(n))+3int_0^1 S^2(n)(nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=n^3-1+int_0^1 S^2(n)(3nx^n-1-2S(n))~dxendalign
beginalign=n^3-1+int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dxendalign
So the question becomes:
Prove beginalignI'=int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dx=1.endalign
beginalignI'=int_0^1 frac3nx^n-1(x^n-1)^2(x-1)^2-frac2(x^n-1)^3(x-1)^3~dxendalign
beginalign=int_0^1 frac(x-1)^2left(frac d dx (x^n-1)^3right)-2(x^n-1)^3(x-1)(x-1)^4~dxendalign
beginalign=int_0^1 frac d dx left(frac(x^n-1)^3(x-1)^2right)~dxendalign
beginalign=lim_x to 1 frac(x^n-1)^3(x-1)^2-frac(0^n-1)^3(0-1)^2endalign
beginaligntherefore I'=1.endalign
beginaligntherefore I_n=n^3.endalign
There MUST be other BETTER ways evaluating $I_n$.
Could anyone give me some better solutions? Thanks.
calculus integration definite-integrals
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
6
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41
add a comment |Â
up vote
12
down vote
favorite
up vote
12
down vote
favorite
How to prove
$~~ forall ninmathbbN^+$,
beginalignI_n=int_0^1(1+x+x^2+x^3+cdotcdotcdot+x^n-1)^2 (1+4x+7x^2+cdotcdotcdot+(3n-2)x^n-1)~dx=n^3.endalign
My Try:
Define $displaystyle S(n)=sum_k=0^n-1x^k=1+x+x^2+x^3+cdotcdotcdot+x^n-1=fracx^n-1x-1$. Then,
beginalignfracddxS(n)=S'(n)=1+2x+3x^2+cdotcdotcdot(n-1)x^n-2=sum_k=0^n-1kx^k-1.endalign
Therefore,
beginalignI_n=int_0^1 S^2(n)left(3S'(n+1)-2S(n)right)~dxendalign
beginalign=3int_0^1 S^2(n)S'(n+1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)(S'(n)+nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)~d(S(n))+3int_0^1 S^2(n)(nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=n^3-1+int_0^1 S^2(n)(3nx^n-1-2S(n))~dxendalign
beginalign=n^3-1+int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dxendalign
So the question becomes:
Prove beginalignI'=int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dx=1.endalign
beginalignI'=int_0^1 frac3nx^n-1(x^n-1)^2(x-1)^2-frac2(x^n-1)^3(x-1)^3~dxendalign
beginalign=int_0^1 frac(x-1)^2left(frac d dx (x^n-1)^3right)-2(x^n-1)^3(x-1)(x-1)^4~dxendalign
beginalign=int_0^1 frac d dx left(frac(x^n-1)^3(x-1)^2right)~dxendalign
beginalign=lim_x to 1 frac(x^n-1)^3(x-1)^2-frac(0^n-1)^3(0-1)^2endalign
beginaligntherefore I'=1.endalign
beginaligntherefore I_n=n^3.endalign
There MUST be other BETTER ways evaluating $I_n$.
Could anyone give me some better solutions? Thanks.
calculus integration definite-integrals
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
How to prove
$~~ forall ninmathbbN^+$,
beginalignI_n=int_0^1(1+x+x^2+x^3+cdotcdotcdot+x^n-1)^2 (1+4x+7x^2+cdotcdotcdot+(3n-2)x^n-1)~dx=n^3.endalign
My Try:
Define $displaystyle S(n)=sum_k=0^n-1x^k=1+x+x^2+x^3+cdotcdotcdot+x^n-1=fracx^n-1x-1$. Then,
beginalignfracddxS(n)=S'(n)=1+2x+3x^2+cdotcdotcdot(n-1)x^n-2=sum_k=0^n-1kx^k-1.endalign
Therefore,
beginalignI_n=int_0^1 S^2(n)left(3S'(n+1)-2S(n)right)~dxendalign
beginalign=3int_0^1 S^2(n)S'(n+1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)(S'(n)+nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=3int_0^1 S^2(n)~d(S(n))+3int_0^1 S^2(n)(nx^n-1)~dx-2int_0^1 S^3(n)~dxendalign
beginalign=n^3-1+int_0^1 S^2(n)(3nx^n-1-2S(n))~dxendalign
beginalign=n^3-1+int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dxendalign
So the question becomes:
Prove beginalignI'=int_0^1 left(fracx^n-1x-1right)^2left(3nx^n-1-2cdotfracx^n-1x-1right)~dx=1.endalign
beginalignI'=int_0^1 frac3nx^n-1(x^n-1)^2(x-1)^2-frac2(x^n-1)^3(x-1)^3~dxendalign
beginalign=int_0^1 frac(x-1)^2left(frac d dx (x^n-1)^3right)-2(x^n-1)^3(x-1)(x-1)^4~dxendalign
beginalign=int_0^1 frac d dx left(frac(x^n-1)^3(x-1)^2right)~dxendalign
beginalign=lim_x to 1 frac(x^n-1)^3(x-1)^2-frac(0^n-1)^3(0-1)^2endalign
beginaligntherefore I'=1.endalign
beginaligntherefore I_n=n^3.endalign
There MUST be other BETTER ways evaluating $I_n$.
Could anyone give me some better solutions? Thanks.
calculus integration definite-integrals
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
edited Aug 12 at 8:10
Martin Sleziak
43.6k6113259
43.6k6113259
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
asked Dec 4 '16 at 8:08
Wong Austin
1,035316
1,035316
6
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41
add a comment |Â
6
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41
6
6
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
35
down vote
accepted
First apply the substitution $x = t^3$. Then
beginalign*
I_n
&= int_0^1 (1 + t^3 + cdots + t^3n-3)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) cdot 3t^2 , dt \
&= int_0^1 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) , dt.
endalign*
Now let $u = u(t) = t + t^4 + cdots + t^3n-2$. Then
$$ 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) = 3u^2 fracdudt.$$
Therefore
$$ I_n = left[ u(t)^3 right]_t=0^t=1 = u(1)^3 - u(0)^3 = n^3. $$
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
35
down vote
accepted
First apply the substitution $x = t^3$. Then
beginalign*
I_n
&= int_0^1 (1 + t^3 + cdots + t^3n-3)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) cdot 3t^2 , dt \
&= int_0^1 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) , dt.
endalign*
Now let $u = u(t) = t + t^4 + cdots + t^3n-2$. Then
$$ 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) = 3u^2 fracdudt.$$
Therefore
$$ I_n = left[ u(t)^3 right]_t=0^t=1 = u(1)^3 - u(0)^3 = n^3. $$
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
add a comment |Â
up vote
35
down vote
accepted
First apply the substitution $x = t^3$. Then
beginalign*
I_n
&= int_0^1 (1 + t^3 + cdots + t^3n-3)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) cdot 3t^2 , dt \
&= int_0^1 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) , dt.
endalign*
Now let $u = u(t) = t + t^4 + cdots + t^3n-2$. Then
$$ 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) = 3u^2 fracdudt.$$
Therefore
$$ I_n = left[ u(t)^3 right]_t=0^t=1 = u(1)^3 - u(0)^3 = n^3. $$
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
add a comment |Â
up vote
35
down vote
accepted
up vote
35
down vote
accepted
First apply the substitution $x = t^3$. Then
beginalign*
I_n
&= int_0^1 (1 + t^3 + cdots + t^3n-3)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) cdot 3t^2 , dt \
&= int_0^1 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) , dt.
endalign*
Now let $u = u(t) = t + t^4 + cdots + t^3n-2$. Then
$$ 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) = 3u^2 fracdudt.$$
Therefore
$$ I_n = left[ u(t)^3 right]_t=0^t=1 = u(1)^3 - u(0)^3 = n^3. $$
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
First apply the substitution $x = t^3$. Then
beginalign*
I_n
&= int_0^1 (1 + t^3 + cdots + t^3n-3)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) cdot 3t^2 , dt \
&= int_0^1 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) , dt.
endalign*
Now let $u = u(t) = t + t^4 + cdots + t^3n-2$. Then
$$ 3 (t + t^4 + cdots + t^3n-2)^2 (1 + 4t^3 + cdots + (3n-2)t^3n-3) = 3u^2 fracdudt.$$
Therefore
$$ I_n = left[ u(t)^3 right]_t=0^t=1 = u(1)^3 - u(0)^3 = n^3. $$
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
edited Dec 4 '16 at 8:19
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
answered Dec 4 '16 at 8:14
Sangchul Lee
85.6k12155253
85.6k12155253
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
add a comment |Â
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
3
3
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
In genious! These are still days when some crazy $u$-substitution surprises me.
– Simply Beautiful Art
Dec 4 '16 at 13:21
add a comment |Â
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6
Out of curiosity, where did this integral appear?
– user159517
Dec 4 '16 at 15:41