Mixing
Compute $int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx$
$begingroup$
Compute $$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}implies I = 3int_0^{infty}frac {u}{u^6+1}du=frac 32int_0^infty frac {1}{u^3+1}du=frac 32 int_0^infty frac {1}{(x+1)(x^2+x+1)}dx\implies I=frac 32 int_0^{infty} frac 1{x+1}-frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
integration definite-integrals improper-integrals
edited Jan 24 at 17:44
Larry
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
$endgroup$
add a comment |
$begingroup$
Compute $$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}implies I = 3int_0^{infty}frac {u}{u^6+1}du=frac 32int_0^infty frac {1}{u^3+1}du=frac 32 int_0^infty frac {1}{(x+1)(x^2+x+1)}dx\implies I=frac 32 int_0^{infty} frac 1{x+1}-frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
integration definite-integrals improper-integrals
edited Jan 24 at 17:44
Larry
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
$endgroup$
$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18
add a comment |
$begingroup$
Compute $$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}implies I = 3int_0^{infty}frac {u}{u^6+1}du=frac 32int_0^infty frac {1}{u^3+1}du=frac 32 int_0^infty frac {1}{(x+1)(x^2+x+1)}dx\implies I=frac 32 int_0^{infty} frac 1{x+1}-frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
integration definite-integrals improper-integrals
edited Jan 24 at 17:44
Larry
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
$endgroup$
Compute $$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx$$
My attempt:
$$u=x^{1/3}implies I = 3int_0^{infty}frac {u}{u^6+1}du=frac 32int_0^infty frac {1}{u^3+1}du=frac 32 int_0^infty frac {1}{(x+1)(x^2+x+1)}dx\implies I=frac 32 int_0^{infty} frac 1{x+1}-frac x{x^2+x+1}dx$$ And I'm stuck here, what can I do from here?
integration definite-integrals improper-integrals
integration definite-integrals improper-integrals
edited Jan 24 at 17:44
Larry
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
edited Jan 24 at 17:44
Larry
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
edited Jan 24 at 17:44
Larry
2,53531131
edited Jan 24 at 17:44
Larry
2,53531131
edited Jan 24 at 17:44
Larry
2,53531131
2,53531131
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
asked Jan 24 at 17:37
C. CristiC. Cristi
1,634218
1,634218
$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18
add a comment |
$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18
$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18
$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18
add a comment |
5 Answers
5
active
oldest
votes
$begingroup$
Hint. Taking from your last step,
$$intfrac x{x^2+x+1}dx=frac{1}{2}intfrac{D(x^2+x+1)}{x^2+x+1}dx-frac{1}{2}intfrac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
|
show 1 more comment
$begingroup$
As an alternative approach:$$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx,overset{large x^{2/3}=u}=,frac32 int_0^infty frac{1}{u^3+1}du$$
We will substitute $displaystyle{u=frac{1-t}{1+t}Rightarrow du=-frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$Rightarrow I=frac32 int_{-1}^1 frac{t+1}{3t^2+1}dt=frac32 cdot 2int_0^1 frac{dt}{3t^2+1}dt=3cdot frac{1}{sqrt 3}arctan(sqrt 3 t)bigg|_0^1=frac{pi}{sqrt 3}$$
answered Jan 24 at 17:42
ZackyZacky
7,86011061
$endgroup$
add a comment |
$begingroup$
Also using the Beta function, but in a different way.
Recall that
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt$$
Using the substitution $w=frac{1-t}t$, we see that
$$mathrm{B}(a,b)=int_0^inftyfrac{w^{b-1}}{(1+w)^{a+b}}mathrm dw$$
Then using the sub $w=u^2$,
$$mathrm{B}(a,b)=2int_0^inftyfrac{u^{2b-1}}{(1+u^2)^{a+b}}mathrm du$$
So setting $2b-1=-1/3$, and $a+b=1$, we see that
$$int_0^infty frac{mathrm dx}{x^{1/3}(1+x^2)}=frac12mathrm{B}(1/3,2/3)$$
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
$endgroup$
add a comment |
$begingroup$
Quickest way would be to let $xmapstotan x$ and use the beta function$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}mathrm dx,tan^{-1/3}x\ & =intlimits_0^{pi/2}mathrm dx,sin^{-1/3}xcos^{1/3}x\ & =frac 12operatorname{B}left(frac 13,frac 23right)end{align*}$$A direct application of Euler’s Reflection formula gives$$intlimits_0^{infty}frac {mathrm dx}{x^{1/3}(1+x^2)}color{blue}{=frac {pi}{sqrt3}}$$
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
$endgroup$
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
add a comment |
$begingroup$
Yet another approach.
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &overset{large x mapsto x^{3/2}}=frac{3}{2} int_0^infty frac{dx}{1 + x^3}\
&= frac{3}{2} int_0^1 frac{dx}{1 + x^3} + frac{3}{2} int_1^infty frac{dx}{1 + x^3}.
end{align}
Enforcing a substitution of $x mapsto 1/x$ in the second of the integrals leads to
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &= frac{3}{2} int_0^1 frac{1 + x}{1 + x^3} , dx\
&= frac{3}{2} int_0^1 frac{dx}{x^2 - x + 1}\
&= frac{3}{2} int_0^1 frac{dx}{left (x - frac{1}{2} right )^2 + frac{3}{4}}\
&= sqrt{3} left [tan^{-1} left (frac{2x - 1}{sqrt{3}} right ) right ]_0^1\
&= frac{pi}{sqrt{3}}.
end{align}
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
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add a comment |
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5 Answers
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active
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5 Answers
5
active
oldest
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active
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active
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$begingroup$
Hint. Taking from your last step,
$$intfrac x{x^2+x+1}dx=frac{1}{2}intfrac{D(x^2+x+1)}{x^2+x+1}dx-frac{1}{2}intfrac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
|
show 1 more comment
$begingroup$
Hint. Taking from your last step,
$$intfrac x{x^2+x+1}dx=frac{1}{2}intfrac{D(x^2+x+1)}{x^2+x+1}dx-frac{1}{2}intfrac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
$endgroup$
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
|
show 1 more comment
$begingroup$
Hint. Taking from your last step,
$$intfrac x{x^2+x+1}dx=frac{1}{2}intfrac{D(x^2+x+1)}{x^2+x+1}dx-frac{1}{2}intfrac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
$endgroup$
Hint. Taking from your last step,
$$intfrac x{x^2+x+1}dx=frac{1}{2}intfrac{D(x^2+x+1)}{x^2+x+1}dx-frac{1}{2}intfrac{1}{(x+1/2)^2+3/4}dx.$$
Can you take it from here?
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
edited Jan 24 at 17:45
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
answered Jan 24 at 17:41
Robert ZRobert Z
101k1069142
101k1069142
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
|
show 1 more comment
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Okay so you added and $x+1$ and subtracted it? that's nice but after the calculation as the primitive I get : $$ln (sqrt{ frac {(x+1^3)}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– C. Cristi
Jan 24 at 17:45
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
Yes. Now that you have the primitive evaluate it at 0 and the limit at $+infty$ and take the difference. The primitive should be $3$ multiplied by $$ln (sqrt{ frac {(x+1)^2}{x^2+x+1} }) + frac 1{sqrt{3}} arctan (frac {(2x+1)}{sqrt{3}})$$
$endgroup$
– Robert Z
Jan 24 at 17:53
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
What if there was actually $int_0^infty frac 1{x^{2/3}(1+x^3)}dx$ What substitution would you use?
$endgroup$
– C. Cristi
Jan 24 at 18:04
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
I would try again $t=x^{1/3}$, but don't think that the primitive is "easy" here. Maybe it's better to change the approach and use complex analysis.
$endgroup$
– Robert Z
Jan 24 at 18:09
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
$begingroup$
Already tried $x^{1/3}$ but it's clearly a lot harder, what kind of complex analysis?
$endgroup$
– C. Cristi
Jan 24 at 18:11
|
show 1 more comment
$begingroup$
As an alternative approach:$$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx,overset{large x^{2/3}=u}=,frac32 int_0^infty frac{1}{u^3+1}du$$
We will substitute $displaystyle{u=frac{1-t}{1+t}Rightarrow du=-frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$Rightarrow I=frac32 int_{-1}^1 frac{t+1}{3t^2+1}dt=frac32 cdot 2int_0^1 frac{dt}{3t^2+1}dt=3cdot frac{1}{sqrt 3}arctan(sqrt 3 t)bigg|_0^1=frac{pi}{sqrt 3}$$
answered Jan 24 at 17:42
ZackyZacky
7,86011061
$endgroup$
add a comment |
$begingroup$
As an alternative approach:$$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx,overset{large x^{2/3}=u}=,frac32 int_0^infty frac{1}{u^3+1}du$$
We will substitute $displaystyle{u=frac{1-t}{1+t}Rightarrow du=-frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$Rightarrow I=frac32 int_{-1}^1 frac{t+1}{3t^2+1}dt=frac32 cdot 2int_0^1 frac{dt}{3t^2+1}dt=3cdot frac{1}{sqrt 3}arctan(sqrt 3 t)bigg|_0^1=frac{pi}{sqrt 3}$$
answered Jan 24 at 17:42
ZackyZacky
7,86011061
$endgroup$
add a comment |
$begingroup$
As an alternative approach:$$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx,overset{large x^{2/3}=u}=,frac32 int_0^infty frac{1}{u^3+1}du$$
We will substitute $displaystyle{u=frac{1-t}{1+t}Rightarrow du=-frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$Rightarrow I=frac32 int_{-1}^1 frac{t+1}{3t^2+1}dt=frac32 cdot 2int_0^1 frac{dt}{3t^2+1}dt=3cdot frac{1}{sqrt 3}arctan(sqrt 3 t)bigg|_0^1=frac{pi}{sqrt 3}$$
answered Jan 24 at 17:42
ZackyZacky
7,86011061
$endgroup$
As an alternative approach:$$I=int_0^{infty} frac {1}{x^{1/3}(1+x^2)}dx,overset{large x^{2/3}=u}=,frac32 int_0^infty frac{1}{u^3+1}du$$
We will substitute $displaystyle{u=frac{1-t}{1+t}Rightarrow du=-frac{2}{(1+t)^2}dt}$.
The reason behind it is that $(1-t)^3+(1+t)^3=2(3t^2+1)$, thus we get rid of the third powers.
$$Rightarrow I=frac32 int_{-1}^1 frac{t+1}{3t^2+1}dt=frac32 cdot 2int_0^1 frac{dt}{3t^2+1}dt=3cdot frac{1}{sqrt 3}arctan(sqrt 3 t)bigg|_0^1=frac{pi}{sqrt 3}$$
answered Jan 24 at 17:42
ZackyZacky
7,86011061
edited Jan 24 at 17:53
answered Jan 24 at 17:42
ZackyZacky
7,86011061
answered Jan 24 at 17:42
ZackyZacky
7,86011061
answered Jan 24 at 17:42
ZackyZacky
7,86011061
7,86011061
add a comment |
add a comment |
$begingroup$
Also using the Beta function, but in a different way.
Recall that
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt$$
Using the substitution $w=frac{1-t}t$, we see that
$$mathrm{B}(a,b)=int_0^inftyfrac{w^{b-1}}{(1+w)^{a+b}}mathrm dw$$
Then using the sub $w=u^2$,
$$mathrm{B}(a,b)=2int_0^inftyfrac{u^{2b-1}}{(1+u^2)^{a+b}}mathrm du$$
So setting $2b-1=-1/3$, and $a+b=1$, we see that
$$int_0^infty frac{mathrm dx}{x^{1/3}(1+x^2)}=frac12mathrm{B}(1/3,2/3)$$
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
$endgroup$
add a comment |
$begingroup$
Also using the Beta function, but in a different way.
Recall that
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt$$
Using the substitution $w=frac{1-t}t$, we see that
$$mathrm{B}(a,b)=int_0^inftyfrac{w^{b-1}}{(1+w)^{a+b}}mathrm dw$$
Then using the sub $w=u^2$,
$$mathrm{B}(a,b)=2int_0^inftyfrac{u^{2b-1}}{(1+u^2)^{a+b}}mathrm du$$
So setting $2b-1=-1/3$, and $a+b=1$, we see that
$$int_0^infty frac{mathrm dx}{x^{1/3}(1+x^2)}=frac12mathrm{B}(1/3,2/3)$$
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
$endgroup$
add a comment |
$begingroup$
Also using the Beta function, but in a different way.
Recall that
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt$$
Using the substitution $w=frac{1-t}t$, we see that
$$mathrm{B}(a,b)=int_0^inftyfrac{w^{b-1}}{(1+w)^{a+b}}mathrm dw$$
Then using the sub $w=u^2$,
$$mathrm{B}(a,b)=2int_0^inftyfrac{u^{2b-1}}{(1+u^2)^{a+b}}mathrm du$$
So setting $2b-1=-1/3$, and $a+b=1$, we see that
$$int_0^infty frac{mathrm dx}{x^{1/3}(1+x^2)}=frac12mathrm{B}(1/3,2/3)$$
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
$endgroup$
Also using the Beta function, but in a different way.
Recall that
$$mathrm{B}(a,b)=int_0^1t^{a-1}(1-t)^{b-1}mathrm dt$$
Using the substitution $w=frac{1-t}t$, we see that
$$mathrm{B}(a,b)=int_0^inftyfrac{w^{b-1}}{(1+w)^{a+b}}mathrm dw$$
Then using the sub $w=u^2$,
$$mathrm{B}(a,b)=2int_0^inftyfrac{u^{2b-1}}{(1+u^2)^{a+b}}mathrm du$$
So setting $2b-1=-1/3$, and $a+b=1$, we see that
$$int_0^infty frac{mathrm dx}{x^{1/3}(1+x^2)}=frac12mathrm{B}(1/3,2/3)$$
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
answered Jan 24 at 19:47
clathratusclathratus
5,0351338
5,0351338
add a comment |
add a comment |
$begingroup$
Quickest way would be to let $xmapstotan x$ and use the beta function$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}mathrm dx,tan^{-1/3}x\ & =intlimits_0^{pi/2}mathrm dx,sin^{-1/3}xcos^{1/3}x\ & =frac 12operatorname{B}left(frac 13,frac 23right)end{align*}$$A direct application of Euler’s Reflection formula gives$$intlimits_0^{infty}frac {mathrm dx}{x^{1/3}(1+x^2)}color{blue}{=frac {pi}{sqrt3}}$$
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
$endgroup$
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
add a comment |
$begingroup$
Quickest way would be to let $xmapstotan x$ and use the beta function$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}mathrm dx,tan^{-1/3}x\ & =intlimits_0^{pi/2}mathrm dx,sin^{-1/3}xcos^{1/3}x\ & =frac 12operatorname{B}left(frac 13,frac 23right)end{align*}$$A direct application of Euler’s Reflection formula gives$$intlimits_0^{infty}frac {mathrm dx}{x^{1/3}(1+x^2)}color{blue}{=frac {pi}{sqrt3}}$$
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
$endgroup$
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
add a comment |
$begingroup$
Quickest way would be to let $xmapstotan x$ and use the beta function$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}mathrm dx,tan^{-1/3}x\ & =intlimits_0^{pi/2}mathrm dx,sin^{-1/3}xcos^{1/3}x\ & =frac 12operatorname{B}left(frac 13,frac 23right)end{align*}$$A direct application of Euler’s Reflection formula gives$$intlimits_0^{infty}frac {mathrm dx}{x^{1/3}(1+x^2)}color{blue}{=frac {pi}{sqrt3}}$$
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
$endgroup$
Quickest way would be to let $xmapstotan x$ and use the beta function$$begin{align*}mathfrak{I} & =intlimits_0^{pi/2}mathrm dx,tan^{-1/3}x\ & =intlimits_0^{pi/2}mathrm dx,sin^{-1/3}xcos^{1/3}x\ & =frac 12operatorname{B}left(frac 13,frac 23right)end{align*}$$A direct application of Euler’s Reflection formula gives$$intlimits_0^{infty}frac {mathrm dx}{x^{1/3}(1+x^2)}color{blue}{=frac {pi}{sqrt3}}$$
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
answered Jan 24 at 17:50
Frank W.Frank W.
3,8101321
3,8101321
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
add a comment |
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
How do you get $B(frac 13, frac 23)$ and why not $B(frac {-1}6, frac 16)?$
$endgroup$
– C. Cristi
Jan 24 at 18:02
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
$begingroup$
@C.Cristi Because$$frac {1}2left(frac 13+1right)=frac 12frac 43=frac 23$$And similarly...
$endgroup$
– Frank W.
Jan 24 at 18:16
add a comment |
$begingroup$
Yet another approach.
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &overset{large x mapsto x^{3/2}}=frac{3}{2} int_0^infty frac{dx}{1 + x^3}\
&= frac{3}{2} int_0^1 frac{dx}{1 + x^3} + frac{3}{2} int_1^infty frac{dx}{1 + x^3}.
end{align}
Enforcing a substitution of $x mapsto 1/x$ in the second of the integrals leads to
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &= frac{3}{2} int_0^1 frac{1 + x}{1 + x^3} , dx\
&= frac{3}{2} int_0^1 frac{dx}{x^2 - x + 1}\
&= frac{3}{2} int_0^1 frac{dx}{left (x - frac{1}{2} right )^2 + frac{3}{4}}\
&= sqrt{3} left [tan^{-1} left (frac{2x - 1}{sqrt{3}} right ) right ]_0^1\
&= frac{pi}{sqrt{3}}.
end{align}
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
$endgroup$
add a comment |
$begingroup$
Yet another approach.
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &overset{large x mapsto x^{3/2}}=frac{3}{2} int_0^infty frac{dx}{1 + x^3}\
&= frac{3}{2} int_0^1 frac{dx}{1 + x^3} + frac{3}{2} int_1^infty frac{dx}{1 + x^3}.
end{align}
Enforcing a substitution of $x mapsto 1/x$ in the second of the integrals leads to
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &= frac{3}{2} int_0^1 frac{1 + x}{1 + x^3} , dx\
&= frac{3}{2} int_0^1 frac{dx}{x^2 - x + 1}\
&= frac{3}{2} int_0^1 frac{dx}{left (x - frac{1}{2} right )^2 + frac{3}{4}}\
&= sqrt{3} left [tan^{-1} left (frac{2x - 1}{sqrt{3}} right ) right ]_0^1\
&= frac{pi}{sqrt{3}}.
end{align}
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
$endgroup$
add a comment |
$begingroup$
Yet another approach.
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &overset{large x mapsto x^{3/2}}=frac{3}{2} int_0^infty frac{dx}{1 + x^3}\
&= frac{3}{2} int_0^1 frac{dx}{1 + x^3} + frac{3}{2} int_1^infty frac{dx}{1 + x^3}.
end{align}
Enforcing a substitution of $x mapsto 1/x$ in the second of the integrals leads to
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &= frac{3}{2} int_0^1 frac{1 + x}{1 + x^3} , dx\
&= frac{3}{2} int_0^1 frac{dx}{x^2 - x + 1}\
&= frac{3}{2} int_0^1 frac{dx}{left (x - frac{1}{2} right )^2 + frac{3}{4}}\
&= sqrt{3} left [tan^{-1} left (frac{2x - 1}{sqrt{3}} right ) right ]_0^1\
&= frac{pi}{sqrt{3}}.
end{align}
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
$endgroup$
Yet another approach.
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &overset{large x mapsto x^{3/2}}=frac{3}{2} int_0^infty frac{dx}{1 + x^3}\
&= frac{3}{2} int_0^1 frac{dx}{1 + x^3} + frac{3}{2} int_1^infty frac{dx}{1 + x^3}.
end{align}
Enforcing a substitution of $x mapsto 1/x$ in the second of the integrals leads to
begin{align}
int_0^infty frac{dx}{sqrt[3]{x} (1 + x^2)} &= frac{3}{2} int_0^1 frac{1 + x}{1 + x^3} , dx\
&= frac{3}{2} int_0^1 frac{dx}{x^2 - x + 1}\
&= frac{3}{2} int_0^1 frac{dx}{left (x - frac{1}{2} right )^2 + frac{3}{4}}\
&= sqrt{3} left [tan^{-1} left (frac{2x - 1}{sqrt{3}} right ) right ]_0^1\
&= frac{pi}{sqrt{3}}.
end{align}
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
edited Jan 25 at 13:07
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
answered Jan 25 at 6:38
omegadotomegadot
6,4072829
6,4072829
add a comment |
add a comment |
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$begingroup$
Note that with the substitution $xmapsto sqrt{x}$, we see that $$int_0^infty frac{1}{x^{1/3}(1+x^2)},dx=frac12int_0^infty frac{x^{-2/3}}{1+x},dx=frac12B(1/3,2/3)=frac12Gamma(1/3)Gamma(2/3)=frac{pi}{2sin(pi/3)}=frac{pi}{sqrt 3}$$
$endgroup$
– Mark Viola
Jan 24 at 21:18