Mixing
contour integral logarithm
$begingroup$
When calculating integrals like $int_{0}^{infty} R(x)log(x) dx$ with R(x)=P(x)/Q(x) a rational function i use the keyhole contour as in the example 4 of this link https://en.wikipedia.org/wiki/Contour_integration, with the argument of the logarithm between o and 2$pi$.
Now everything is fine if Q doesn't have non negative zeros, but if it has i haven't found anything on the internet. I think i should change the argument and choose the one between -$pi$ and $pi$, but in that case i think i end up with $int_{0}^{infty} R(-x)(logx+ i{pi})^2 dx$ - $int_{0}^{infty} R(-x)(logx- i{pi})^2 dx$ .
Now that R(-x) is my mistake, it is actually R(x) and everything is fine or we can use this type of contour only for special cases like if R is even or odd?
And if Q has both positive and negative real zeros how should i do? I was thinking to use the keyhole contour with argumentof the log between 0 and 2 $pi$ and then "draw" a full circle around the positive real singularity that we call x and using the small circle lemma i have 2$pi$il with
l=$lim_(z->x)(z-x)(R(z)(logz)^2$ is it ok?
Finally i think 1 is a simple zero for the complex log isn't it?
definite-integrals logarithms residue-calculus
asked Jan 7 at 3:15
the crazythe crazy
63
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add a comment |
$begingroup$
When calculating integrals like $int_{0}^{infty} R(x)log(x) dx$ with R(x)=P(x)/Q(x) a rational function i use the keyhole contour as in the example 4 of this link https://en.wikipedia.org/wiki/Contour_integration, with the argument of the logarithm between o and 2$pi$.
Now everything is fine if Q doesn't have non negative zeros, but if it has i haven't found anything on the internet. I think i should change the argument and choose the one between -$pi$ and $pi$, but in that case i think i end up with $int_{0}^{infty} R(-x)(logx+ i{pi})^2 dx$ - $int_{0}^{infty} R(-x)(logx- i{pi})^2 dx$ .
Now that R(-x) is my mistake, it is actually R(x) and everything is fine or we can use this type of contour only for special cases like if R is even or odd?
And if Q has both positive and negative real zeros how should i do? I was thinking to use the keyhole contour with argumentof the log between 0 and 2 $pi$ and then "draw" a full circle around the positive real singularity that we call x and using the small circle lemma i have 2$pi$il with
l=$lim_(z->x)(z-x)(R(z)(logz)^2$ is it ok?
Finally i think 1 is a simple zero for the complex log isn't it?
definite-integrals logarithms residue-calculus
asked Jan 7 at 3:15
the crazythe crazy
63
$endgroup$
add a comment |
$begingroup$
When calculating integrals like $int_{0}^{infty} R(x)log(x) dx$ with R(x)=P(x)/Q(x) a rational function i use the keyhole contour as in the example 4 of this link https://en.wikipedia.org/wiki/Contour_integration, with the argument of the logarithm between o and 2$pi$.
Now everything is fine if Q doesn't have non negative zeros, but if it has i haven't found anything on the internet. I think i should change the argument and choose the one between -$pi$ and $pi$, but in that case i think i end up with $int_{0}^{infty} R(-x)(logx+ i{pi})^2 dx$ - $int_{0}^{infty} R(-x)(logx- i{pi})^2 dx$ .
Now that R(-x) is my mistake, it is actually R(x) and everything is fine or we can use this type of contour only for special cases like if R is even or odd?
And if Q has both positive and negative real zeros how should i do? I was thinking to use the keyhole contour with argumentof the log between 0 and 2 $pi$ and then "draw" a full circle around the positive real singularity that we call x and using the small circle lemma i have 2$pi$il with
l=$lim_(z->x)(z-x)(R(z)(logz)^2$ is it ok?
Finally i think 1 is a simple zero for the complex log isn't it?
definite-integrals logarithms residue-calculus
asked Jan 7 at 3:15
the crazythe crazy
63
$endgroup$
When calculating integrals like $int_{0}^{infty} R(x)log(x) dx$ with R(x)=P(x)/Q(x) a rational function i use the keyhole contour as in the example 4 of this link https://en.wikipedia.org/wiki/Contour_integration, with the argument of the logarithm between o and 2$pi$.
Now everything is fine if Q doesn't have non negative zeros, but if it has i haven't found anything on the internet. I think i should change the argument and choose the one between -$pi$ and $pi$, but in that case i think i end up with $int_{0}^{infty} R(-x)(logx+ i{pi})^2 dx$ - $int_{0}^{infty} R(-x)(logx- i{pi})^2 dx$ .
Now that R(-x) is my mistake, it is actually R(x) and everything is fine or we can use this type of contour only for special cases like if R is even or odd?
And if Q has both positive and negative real zeros how should i do? I was thinking to use the keyhole contour with argumentof the log between 0 and 2 $pi$ and then "draw" a full circle around the positive real singularity that we call x and using the small circle lemma i have 2$pi$il with
l=$lim_(z->x)(z-x)(R(z)(logz)^2$ is it ok?
Finally i think 1 is a simple zero for the complex log isn't it?
definite-integrals logarithms residue-calculus
definite-integrals logarithms residue-calculus
asked Jan 7 at 3:15
the crazythe crazy
63
asked Jan 7 at 3:15
the crazythe crazy
63
asked Jan 7 at 3:15
the crazythe crazy
63
asked Jan 7 at 3:15
the crazythe crazy
63
asked Jan 7 at 3:15
the crazythe crazy
63
63
add a comment |
add a comment |
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