Why does this integral equal $0$?
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In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$
integration definite-integrals symmetry
asked Nov 20 at 0:42
Frpzzd
19.8k638101
add a comment |
up vote
11
down vote
favorite
In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$
integration definite-integrals symmetry
asked Nov 20 at 0:42
Frpzzd
19.8k638101
I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47
add a comment |
up vote
11
down vote
favorite
up vote
11
down vote
favorite
In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$
integration definite-integrals symmetry
asked Nov 20 at 0:42
Frpzzd
19.8k638101
In this question, the OP poses the following definite integral, which just happens to vanish:
$$int_1^sqrt2 frac{1}{x}lnbigg(frac{2-2x^2+x^4}{2x-2x^2+x^3}bigg)dx=0$$
As noticed by one commenter to the question, the only zero of the integrand is at $x=sqrt[3]{2}$, meaning that the integral of the integrand from $x=1$ to $x=sqrt[3]{2}$ is the additive inverse of the integral of the integrand from $x=sqrt[3]{2}$ to $x=sqrt{2}$.
This suggests some sort of symmetry obtainable by a substitution, but I cannot find an appropriate substitution or cancellation. It seems like the answer should be much simpler than those posted to the linked question.
Any ideas?
EDIT: I believe that this more general integral also vanishes:
$$int_1^{sqrt{t}}frac{1}{x}lnbigg(frac{t-sx^2+x^4}{tx-sx^2+x^3}bigg)dx=0$$
integration definite-integrals symmetry
integration definite-integrals symmetry
asked Nov 20 at 0:42
Frpzzd
19.8k638101
asked Nov 20 at 0:42
Frpzzd
19.8k638101
edited 2 days ago
asked Nov 20 at 0:42
Frpzzd
19.8k638101
asked Nov 20 at 0:42
Frpzzd
19.8k638101
asked Nov 20 at 0:42
Frpzzd
19.8k638101
19.8k638101
I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47
add a comment |
I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47
I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47
add a comment |
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I wouldn't say the integrand is symmetric around $x=sqrt[3]{2}$ since $sqrt[3]{2}$ is not the midpoint of $1$ and $sqrt{2}$.
– Rócherz
Nov 20 at 0:46
@Rócherz Of course it isn't symmetric about $sqrt[3]{2}$; I simply mean that perhaps an appropriate substitution might lead to a cancellation.
– Frpzzd
Nov 20 at 0:47