Mixing
Evaluate $int_{0}^{1}frac{1+x+x^2}{1+x+x^2+x^3+x^4}dx$
$begingroup$
Evaluate $$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=frac{1-x^5}{1-x}$$
So we get:
$$I=int_{0}^{1}frac{1-x^3}{1-x^5}dx$$
$$I=1+int_{0}^{1}frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
integration definite-integrals
edited Jan 28 at 14:36
Zacky
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
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add a comment |
$begingroup$
Evaluate $$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=frac{1-x^5}{1-x}$$
So we get:
$$I=int_{0}^{1}frac{1-x^3}{1-x^5}dx$$
$$I=1+int_{0}^{1}frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
integration definite-integrals
edited Jan 28 at 14:36
Zacky
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
$endgroup$
add a comment |
$begingroup$
Evaluate $$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=frac{1-x^5}{1-x}$$
So we get:
$$I=int_{0}^{1}frac{1-x^3}{1-x^5}dx$$
$$I=1+int_{0}^{1}frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
integration definite-integrals
edited Jan 28 at 14:36
Zacky
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
$endgroup$
Evaluate $$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx$$
My try:
We have:
$$1+x+x^2=frac{1-x^3}{1-x}$$
$$1+x+x^2+x^3+x^4=frac{1-x^5}{1-x}$$
So we get:
$$I=int_{0}^{1}frac{1-x^3}{1-x^5}dx$$
$$I=1+int_{0}^{1}frac{x^3(x^2-1)}{x^5-1}dx$$
Any idea from here?
integration definite-integrals
integration definite-integrals
edited Jan 28 at 14:36
Zacky
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
edited Jan 28 at 14:36
Zacky
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
edited Jan 28 at 14:36
Zacky
7,76511062
edited Jan 28 at 14:36
Zacky
7,76511062
edited Jan 28 at 14:36
Zacky
7,76511062
7,76511062
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
asked Jan 28 at 14:23
Umesh shankarUmesh shankar
3,07931220
3,07931220
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
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Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=left(x^2+frac{1+sqrt5}2x+1right)left(x^2+frac{1-sqrt5}2x+1right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$left(sqrt5+1right)left(x^2+frac{1-sqrt5}2x+1right)+left(sqrt5-1right)left(x^2+frac{1+sqrt5}2x+1right)=2sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$int_0^1frac{dx}{x^2+2ax+1}=int_0^1frac{dx}{(x+a)^2+1-a^2}=left.frac1{sqrt{1-a^2}}arctanfrac{x+a}{sqrt{1-a^2}}right|_0^1.$$
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
add a comment |
$begingroup$
This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=int_0^1 frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $displaystyle{frac{1}{1-x^z}=sum_{n=0}^infty x^{nz}, , |x|<1 } $. Applying it here: $$I=int_0^1 (1-x^3) left(sum_{n=0}^infty x^{5n}right)dx=sum_{n=0}^infty int_0^1 left(x^{5n}-x^{5n+3}right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=left(frac{x^{5n+1}} {5n+1} - frac{x^{5n+4}} {5n+4}right) bigg|_0^1 =sum_{n=0}^inftyleft(frac{1} {5n+1} - frac{1 } {5n+4}right) =frac15 sum_{n=0}^inftyleft(frac{1} {n+frac15} - frac{1 } {n+frac45}right) $$
We have by the series formula of digamma function that:
$$psi(z)-psi(s)=left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{z+n}right)-left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{s+n}right)=sum_{n=1}^inftyleft(frac{1}{s+n}-frac{1}{z+n}right) $$ So $displaystyle{ I=frac15left(psileft(frac45right) - psileft(frac15right)right) } $. Also using the reflection formula:$$psi(1-z)-psi(z)=picot(pi z)$$ We have: $$I=frac{pi} {5}cotleft(frac{pi}{5}right)=frac{pi} {5}sqrt{1 +frac{2} {sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=frac{1-t}{1+t}rightarrow dx=-frac{2}{(1+t)^2}dt$$
$$Rightarrow I=2int_0^1 frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=frac{sqrt 5+1}{sqrt 5}int_0^1 frac{dx}{x^2+2sqrt 5+5}+frac{sqrt 5-1}{sqrt 5}int_0^1 frac{dx}{x^2-2sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
answered Jan 28 at 14:34
ZackyZacky
7,76511062
$endgroup$
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
$begingroup$
Elementary method is quite nice for me, thanks
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– Umesh shankar
Jan 29 at 18:49
add a comment |
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The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=int_0^1frac{1-x^3}{1-x^5}mathrm dx=int_0^1sum_{n=0}^infty x^{5n}(1-x^3)mathrm dx=sum_{n=0}^inftyint_0^1x^{5n}-x^{5n+3}mathrm dx\=sum_{n=0}^inftyleft[frac{x^{5n+1}}{5n+1}-frac{x^{5n+4}}{5n+4}right]_0^1
=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=1-frac14+frac16-frac19+cdots=1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$picot(pi z)~=~frac1z+sum_{n=1}^{infty}frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
begin{align}
1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]&=1-sum_{n=1}^inftyfrac2{25n^2-1}\
&=1+frac15left[sum_{n=1}^inftyfrac{2frac15}{left(frac15right)^2-n^2}right]\
&=1+frac15left[picotleft(fracpi5right)-5right]\
&=fracpi5cotleft(fracpi5right)
end{align}
$$therefore~I=int_0^1frac{1+x+x^2}{1+x+x^2+x^3+x^4}mathrm dx=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=fracpi5sqrt{1+frac2{sqrt 5}}$$
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
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add a comment |
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Your integral $$ int_{0}^{1}frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-gamma+int_{0}^{1}frac{1-t^s}{1-t}dt=psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
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Is this an actual solution or just a first thought?
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– mrtaurho
Jan 28 at 15:02
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
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– Zacky
Jan 28 at 15:08
2
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Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
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– Zacky
Jan 28 at 15:14
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4 Answers
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4 Answers
4
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oldest
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active
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$begingroup$
Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=left(x^2+frac{1+sqrt5}2x+1right)left(x^2+frac{1-sqrt5}2x+1right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$left(sqrt5+1right)left(x^2+frac{1-sqrt5}2x+1right)+left(sqrt5-1right)left(x^2+frac{1+sqrt5}2x+1right)=2sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$int_0^1frac{dx}{x^2+2ax+1}=int_0^1frac{dx}{(x+a)^2+1-a^2}=left.frac1{sqrt{1-a^2}}arctanfrac{x+a}{sqrt{1-a^2}}right|_0^1.$$
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
add a comment |
$begingroup$
Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=left(x^2+frac{1+sqrt5}2x+1right)left(x^2+frac{1-sqrt5}2x+1right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$left(sqrt5+1right)left(x^2+frac{1-sqrt5}2x+1right)+left(sqrt5-1right)left(x^2+frac{1+sqrt5}2x+1right)=2sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$int_0^1frac{dx}{x^2+2ax+1}=int_0^1frac{dx}{(x+a)^2+1-a^2}=left.frac1{sqrt{1-a^2}}arctanfrac{x+a}{sqrt{1-a^2}}right|_0^1.$$
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
$endgroup$
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
add a comment |
$begingroup$
Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=left(x^2+frac{1+sqrt5}2x+1right)left(x^2+frac{1-sqrt5}2x+1right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$left(sqrt5+1right)left(x^2+frac{1-sqrt5}2x+1right)+left(sqrt5-1right)left(x^2+frac{1+sqrt5}2x+1right)=2sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$int_0^1frac{dx}{x^2+2ax+1}=int_0^1frac{dx}{(x+a)^2+1-a^2}=left.frac1{sqrt{1-a^2}}arctanfrac{x+a}{sqrt{1-a^2}}right|_0^1.$$
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
$endgroup$
Hint:
As the roots of the denominator are the complex fifth roots of unity, you find the factorization to be
$$1+x+x^2+x^3+x^4=left(x^2+frac{1+sqrt5}2x+1right)left(x^2+frac{1-sqrt5}2x+1right).$$
Then we can find a linear combination to obtain a quadratic polynomial with equal coefficients,
$$left(sqrt5+1right)left(x^2+frac{1-sqrt5}2x+1right)+left(sqrt5-1right)left(x^2+frac{1+sqrt5}2x+1right)=2sqrt5(x^2+x+1)$$ which leads us to the decomposition in simple fractions.
Now,
$$int_0^1frac{dx}{x^2+2ax+1}=int_0^1frac{dx}{(x+a)^2+1-a^2}=left.frac1{sqrt{1-a^2}}arctanfrac{x+a}{sqrt{1-a^2}}right|_0^1.$$
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
edited Jan 28 at 15:39
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
answered Jan 28 at 15:27
Yves DaoustYves Daoust
131k676229
131k676229
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
add a comment |
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
1
1
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
$begingroup$
Excellent solution, i liked it very much
$endgroup$
– Umesh shankar
Jan 29 at 4:50
add a comment |
$begingroup$
This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=int_0^1 frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $displaystyle{frac{1}{1-x^z}=sum_{n=0}^infty x^{nz}, , |x|<1 } $. Applying it here: $$I=int_0^1 (1-x^3) left(sum_{n=0}^infty x^{5n}right)dx=sum_{n=0}^infty int_0^1 left(x^{5n}-x^{5n+3}right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=left(frac{x^{5n+1}} {5n+1} - frac{x^{5n+4}} {5n+4}right) bigg|_0^1 =sum_{n=0}^inftyleft(frac{1} {5n+1} - frac{1 } {5n+4}right) =frac15 sum_{n=0}^inftyleft(frac{1} {n+frac15} - frac{1 } {n+frac45}right) $$
We have by the series formula of digamma function that:
$$psi(z)-psi(s)=left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{z+n}right)-left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{s+n}right)=sum_{n=1}^inftyleft(frac{1}{s+n}-frac{1}{z+n}right) $$ So $displaystyle{ I=frac15left(psileft(frac45right) - psileft(frac15right)right) } $. Also using the reflection formula:$$psi(1-z)-psi(z)=picot(pi z)$$ We have: $$I=frac{pi} {5}cotleft(frac{pi}{5}right)=frac{pi} {5}sqrt{1 +frac{2} {sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=frac{1-t}{1+t}rightarrow dx=-frac{2}{(1+t)^2}dt$$
$$Rightarrow I=2int_0^1 frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=frac{sqrt 5+1}{sqrt 5}int_0^1 frac{dx}{x^2+2sqrt 5+5}+frac{sqrt 5-1}{sqrt 5}int_0^1 frac{dx}{x^2-2sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
answered Jan 28 at 14:34
ZackyZacky
7,76511062
$endgroup$
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
add a comment |
$begingroup$
This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=int_0^1 frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $displaystyle{frac{1}{1-x^z}=sum_{n=0}^infty x^{nz}, , |x|<1 } $. Applying it here: $$I=int_0^1 (1-x^3) left(sum_{n=0}^infty x^{5n}right)dx=sum_{n=0}^infty int_0^1 left(x^{5n}-x^{5n+3}right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=left(frac{x^{5n+1}} {5n+1} - frac{x^{5n+4}} {5n+4}right) bigg|_0^1 =sum_{n=0}^inftyleft(frac{1} {5n+1} - frac{1 } {5n+4}right) =frac15 sum_{n=0}^inftyleft(frac{1} {n+frac15} - frac{1 } {n+frac45}right) $$
We have by the series formula of digamma function that:
$$psi(z)-psi(s)=left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{z+n}right)-left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{s+n}right)=sum_{n=1}^inftyleft(frac{1}{s+n}-frac{1}{z+n}right) $$ So $displaystyle{ I=frac15left(psileft(frac45right) - psileft(frac15right)right) } $. Also using the reflection formula:$$psi(1-z)-psi(z)=picot(pi z)$$ We have: $$I=frac{pi} {5}cotleft(frac{pi}{5}right)=frac{pi} {5}sqrt{1 +frac{2} {sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=frac{1-t}{1+t}rightarrow dx=-frac{2}{(1+t)^2}dt$$
$$Rightarrow I=2int_0^1 frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=frac{sqrt 5+1}{sqrt 5}int_0^1 frac{dx}{x^2+2sqrt 5+5}+frac{sqrt 5-1}{sqrt 5}int_0^1 frac{dx}{x^2-2sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
answered Jan 28 at 14:34
ZackyZacky
7,76511062
$endgroup$
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
add a comment |
$begingroup$
This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=int_0^1 frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $displaystyle{frac{1}{1-x^z}=sum_{n=0}^infty x^{nz}, , |x|<1 } $. Applying it here: $$I=int_0^1 (1-x^3) left(sum_{n=0}^infty x^{5n}right)dx=sum_{n=0}^infty int_0^1 left(x^{5n}-x^{5n+3}right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=left(frac{x^{5n+1}} {5n+1} - frac{x^{5n+4}} {5n+4}right) bigg|_0^1 =sum_{n=0}^inftyleft(frac{1} {5n+1} - frac{1 } {5n+4}right) =frac15 sum_{n=0}^inftyleft(frac{1} {n+frac15} - frac{1 } {n+frac45}right) $$
We have by the series formula of digamma function that:
$$psi(z)-psi(s)=left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{z+n}right)-left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{s+n}right)=sum_{n=1}^inftyleft(frac{1}{s+n}-frac{1}{z+n}right) $$ So $displaystyle{ I=frac15left(psileft(frac45right) - psileft(frac15right)right) } $. Also using the reflection formula:$$psi(1-z)-psi(z)=picot(pi z)$$ We have: $$I=frac{pi} {5}cotleft(frac{pi}{5}right)=frac{pi} {5}sqrt{1 +frac{2} {sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=frac{1-t}{1+t}rightarrow dx=-frac{2}{(1+t)^2}dt$$
$$Rightarrow I=2int_0^1 frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=frac{sqrt 5+1}{sqrt 5}int_0^1 frac{dx}{x^2+2sqrt 5+5}+frac{sqrt 5-1}{sqrt 5}int_0^1 frac{dx}{x^2-2sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
answered Jan 28 at 14:34
ZackyZacky
7,76511062
$endgroup$
This integral appeared on AoPS a while ago, it might be that it was on MSE too. Anyway I will quote what I did on AoPS.
$$I=int_{0}^{1}frac{(1+x+x^2)}{1+x+x^2+x^3+x^4}dx=int_0^1 frac{1-x^3 }{1-x^5}dx$$ Recall the geometric series: $displaystyle{frac{1}{1-x^z}=sum_{n=0}^infty x^{nz}, , |x|<1 } $. Applying it here: $$I=int_0^1 (1-x^3) left(sum_{n=0}^infty x^{5n}right)dx=sum_{n=0}^infty int_0^1 left(x^{5n}-x^{5n+3}right)dx$$ I don't know how to show that we can swap the sum and the integral here. $$I=left(frac{x^{5n+1}} {5n+1} - frac{x^{5n+4}} {5n+4}right) bigg|_0^1 =sum_{n=0}^inftyleft(frac{1} {5n+1} - frac{1 } {5n+4}right) =frac15 sum_{n=0}^inftyleft(frac{1} {n+frac15} - frac{1 } {n+frac45}right) $$
We have by the series formula of digamma function that:
$$psi(z)-psi(s)=left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{z+n}right)-left(-gamma-1+sum_{n=1}^inftyfrac{1}{n}-frac{1}{s+n}right)=sum_{n=1}^inftyleft(frac{1}{s+n}-frac{1}{z+n}right) $$ So $displaystyle{ I=frac15left(psileft(frac45right) - psileft(frac15right)right) } $. Also using the reflection formula:$$psi(1-z)-psi(z)=picot(pi z)$$ We have: $$I=frac{pi} {5}cotleft(frac{pi}{5}right)=frac{pi} {5}sqrt{1 +frac{2} {sqrt 5}}$$ see also: https://en.m.wikipedia.org/wiki/Digamma_function and http://mathworld.wolfram.com/TrigonometryAnglesPi5.html
New solution and quite elementary. Let:
$$x=frac{1-t}{1+t}rightarrow dx=-frac{2}{(1+t)^2}dt$$
$$Rightarrow I=2int_0^1 frac{x^2+3}{x^4+10x^2+5}dx$$
And by performing partial fractions we get:
$$I=frac{sqrt 5+1}{sqrt 5}int_0^1 frac{dx}{x^2+2sqrt 5+5}+frac{sqrt 5-1}{sqrt 5}int_0^1 frac{dx}{x^2-2sqrt 5+5}$$
And now there are two simple integrals left and some algebra.
answered Jan 28 at 14:34
ZackyZacky
7,76511062
edited Jan 28 at 14:59
answered Jan 28 at 14:34
ZackyZacky
7,76511062
answered Jan 28 at 14:34
ZackyZacky
7,76511062
answered Jan 28 at 14:34
ZackyZacky
7,76511062
7,76511062
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
add a comment |
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
4
4
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
$begingroup$
Really neat use of $$cotfracpi5=sqrt{1+frac2{sqrt5}}$$
$endgroup$
– clathratus
Jan 28 at 16:11
1
1
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
$begingroup$
Elementary method is quite nice for me, thanks
$endgroup$
– Umesh shankar
Jan 29 at 18:49
add a comment |
$begingroup$
The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=int_0^1frac{1-x^3}{1-x^5}mathrm dx=int_0^1sum_{n=0}^infty x^{5n}(1-x^3)mathrm dx=sum_{n=0}^inftyint_0^1x^{5n}-x^{5n+3}mathrm dx\=sum_{n=0}^inftyleft[frac{x^{5n+1}}{5n+1}-frac{x^{5n+4}}{5n+4}right]_0^1
=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=1-frac14+frac16-frac19+cdots=1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$picot(pi z)~=~frac1z+sum_{n=1}^{infty}frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
begin{align}
1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]&=1-sum_{n=1}^inftyfrac2{25n^2-1}\
&=1+frac15left[sum_{n=1}^inftyfrac{2frac15}{left(frac15right)^2-n^2}right]\
&=1+frac15left[picotleft(fracpi5right)-5right]\
&=fracpi5cotleft(fracpi5right)
end{align}
$$therefore~I=int_0^1frac{1+x+x^2}{1+x+x^2+x^3+x^4}mathrm dx=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=fracpi5sqrt{1+frac2{sqrt 5}}$$
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
$endgroup$
add a comment |
$begingroup$
The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=int_0^1frac{1-x^3}{1-x^5}mathrm dx=int_0^1sum_{n=0}^infty x^{5n}(1-x^3)mathrm dx=sum_{n=0}^inftyint_0^1x^{5n}-x^{5n+3}mathrm dx\=sum_{n=0}^inftyleft[frac{x^{5n+1}}{5n+1}-frac{x^{5n+4}}{5n+4}right]_0^1
=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=1-frac14+frac16-frac19+cdots=1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$picot(pi z)~=~frac1z+sum_{n=1}^{infty}frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
begin{align}
1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]&=1-sum_{n=1}^inftyfrac2{25n^2-1}\
&=1+frac15left[sum_{n=1}^inftyfrac{2frac15}{left(frac15right)^2-n^2}right]\
&=1+frac15left[picotleft(fracpi5right)-5right]\
&=fracpi5cotleft(fracpi5right)
end{align}
$$therefore~I=int_0^1frac{1+x+x^2}{1+x+x^2+x^3+x^4}mathrm dx=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=fracpi5sqrt{1+frac2{sqrt 5}}$$
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
$endgroup$
add a comment |
$begingroup$
The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=int_0^1frac{1-x^3}{1-x^5}mathrm dx=int_0^1sum_{n=0}^infty x^{5n}(1-x^3)mathrm dx=sum_{n=0}^inftyint_0^1x^{5n}-x^{5n+3}mathrm dx\=sum_{n=0}^inftyleft[frac{x^{5n+1}}{5n+1}-frac{x^{5n+4}}{5n+4}right]_0^1
=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=1-frac14+frac16-frac19+cdots=1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$picot(pi z)~=~frac1z+sum_{n=1}^{infty}frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
begin{align}
1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]&=1-sum_{n=1}^inftyfrac2{25n^2-1}\
&=1+frac15left[sum_{n=1}^inftyfrac{2frac15}{left(frac15right)^2-n^2}right]\
&=1+frac15left[picotleft(fracpi5right)-5right]\
&=fracpi5cotleft(fracpi5right)
end{align}
$$therefore~I=int_0^1frac{1+x+x^2}{1+x+x^2+x^3+x^4}mathrm dx=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=fracpi5sqrt{1+frac2{sqrt 5}}$$
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
$endgroup$
The reduced form you gave in the end is a good starting point. Since we are within the interval $[0,1]$ we may expand the denominator as its corresponding geometric series, followed by changing the order of summation and integration and finally by termwise integration. Overall this leads us to
$$I=int_0^1frac{1-x^3}{1-x^5}mathrm dx=int_0^1sum_{n=0}^infty x^{5n}(1-x^3)mathrm dx=sum_{n=0}^inftyint_0^1x^{5n}-x^{5n+3}mathrm dx\=sum_{n=0}^inftyleft[frac{x^{5n+1}}{5n+1}-frac{x^{5n+4}}{5n+4}right]_0^1
=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]$$
The difficult task is to find a closed-form for this sum. Of course, one could invoke the Digamma Function as Zacky quoted, but we can also note the following due the absolute convergence of the series
$$I=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=1-frac14+frac16-frac19+cdots=1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]$$
The latter sum falls down quite easy by considering a well-known series expansion of the cotangent function
$$picot(pi z)~=~frac1z+sum_{n=1}^{infty}frac{2z}{z^2-n^2} $$
Thus, we rewrite the original sum as
begin{align}
1+sum_{n=1}^inftyleft[frac{1}{5n+1}-frac{1}{5n-1}right]&=1-sum_{n=1}^inftyfrac2{25n^2-1}\
&=1+frac15left[sum_{n=1}^inftyfrac{2frac15}{left(frac15right)^2-n^2}right]\
&=1+frac15left[picotleft(fracpi5right)-5right]\
&=fracpi5cotleft(fracpi5right)
end{align}
$$therefore~I=int_0^1frac{1+x+x^2}{1+x+x^2+x^3+x^4}mathrm dx=sum_{n=0}^inftyleft[frac{1}{5n+1}-frac{1}{5n+4}right]=fracpi5sqrt{1+frac2{sqrt 5}}$$
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
edited Jan 28 at 15:07
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
answered Jan 28 at 14:55
mrtaurhomrtaurho
6,09271641
6,09271641
add a comment |
add a comment |
$begingroup$
Your integral $$ int_{0}^{1}frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-gamma+int_{0}^{1}frac{1-t^s}{1-t}dt=psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
$endgroup$
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
2
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
add a comment |
$begingroup$
Your integral $$ int_{0}^{1}frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-gamma+int_{0}^{1}frac{1-t^s}{1-t}dt=psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
$endgroup$
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
2
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
add a comment |
$begingroup$
Your integral $$ int_{0}^{1}frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-gamma+int_{0}^{1}frac{1-t^s}{1-t}dt=psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
$endgroup$
Your integral $$ int_{0}^{1}frac{1-x^3}{1-x^5}dx$$ is similar to that of the digamma function $$-gamma+int_{0}^{1}frac{1-t^s}{1-t}dt=psi(s + 1) $$Maybe the change of variable $t=x^5$ could be helpful.
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
answered Jan 28 at 14:36
UrbanmathsUrbanmaths
714
714
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
2
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
add a comment |
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
2
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
$begingroup$
Is this an actual solution or just a first thought?
$endgroup$
– mrtaurho
Jan 28 at 15:02
3
3
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
$begingroup$
This might work indeed, but not that simple. By letting $x^5=t$ we get: $$I=int_0^1 frac{1-x^3}{1-x^5}dx=frac15int_0^1 frac{1-t^{3/5}}{1-t} t^{-4/5}dt$$ Now we need to do a cute trick, add $0 (1-1)$ in the numerator and split into two integrals as: $$I=frac15int_0^1 frac{1-t^{-4/5}cdot t^{3/5}}{1-t}dt-frac15int_0^1 frac{1-t^{-4/5}}{1-t}dt $$ And now apply that definition, $gamma$ gets canceled out of course.
$endgroup$
– Zacky
Jan 28 at 15:08
2
2
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
$begingroup$
Of course there is no surprise that it works, and that we have a difference of digamma functions looking at other answers.
$endgroup$
– Zacky
Jan 28 at 15:14
add a comment |
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