Mixing
Laurent series of an integral with parameter
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To find the Laurent series of function $f(a)$ at point $a=0$
$$
f(a)=int^1_0 fracd xx^2+a^2
$$
one can first do the integral
$$
f(a)=frac1aarctan(1/a)
$$
then expand $arctan(1/a)$ and obtain
$$
f(a)approx fracpi2a-1+O(a^2).
$$
However if one expands the integrand before integration,
one would get divergent integral.
My question is: is there any method helping us expand the function $f(a)$ before performing the integral?
The motivation arises from more complicated cases, when the functions can not be integrated out analytically.
==========================================
Updated: I think I find the solution to this integral. When I expand the function $f(a)$, I have to expand all the integral, not only the integrand
$$
frac1x^2+a^2approx frac1x^2+O(a^2)
$$
which means that the upper limit of integral has also to be changed, i.e.
$$
f(a)approx int^1_a d xleft(frac1x^2+O(a^2)right)
$$
this gives us
$$
f(a)approx frac1315 a-1+O(a^2)
$$
Except the coefficient of $a^-1$, all the other orders are consistent.
divergent-series laurent-series divergence regularization
asked Jul 31 at 13:10
user142288
11
add a comment |Â
up vote
0
down vote
favorite
To find the Laurent series of function $f(a)$ at point $a=0$
$$
f(a)=int^1_0 fracd xx^2+a^2
$$
one can first do the integral
$$
f(a)=frac1aarctan(1/a)
$$
then expand $arctan(1/a)$ and obtain
$$
f(a)approx fracpi2a-1+O(a^2).
$$
However if one expands the integrand before integration,
one would get divergent integral.
My question is: is there any method helping us expand the function $f(a)$ before performing the integral?
The motivation arises from more complicated cases, when the functions can not be integrated out analytically.
==========================================
Updated: I think I find the solution to this integral. When I expand the function $f(a)$, I have to expand all the integral, not only the integrand
$$
frac1x^2+a^2approx frac1x^2+O(a^2)
$$
which means that the upper limit of integral has also to be changed, i.e.
$$
f(a)approx int^1_a d xleft(frac1x^2+O(a^2)right)
$$
this gives us
$$
f(a)approx frac1315 a-1+O(a^2)
$$
Except the coefficient of $a^-1$, all the other orders are consistent.
divergent-series laurent-series divergence regularization
asked Jul 31 at 13:10
user142288
11
Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
To find the Laurent series of function $f(a)$ at point $a=0$
$$
f(a)=int^1_0 fracd xx^2+a^2
$$
one can first do the integral
$$
f(a)=frac1aarctan(1/a)
$$
then expand $arctan(1/a)$ and obtain
$$
f(a)approx fracpi2a-1+O(a^2).
$$
However if one expands the integrand before integration,
one would get divergent integral.
My question is: is there any method helping us expand the function $f(a)$ before performing the integral?
The motivation arises from more complicated cases, when the functions can not be integrated out analytically.
==========================================
Updated: I think I find the solution to this integral. When I expand the function $f(a)$, I have to expand all the integral, not only the integrand
$$
frac1x^2+a^2approx frac1x^2+O(a^2)
$$
which means that the upper limit of integral has also to be changed, i.e.
$$
f(a)approx int^1_a d xleft(frac1x^2+O(a^2)right)
$$
this gives us
$$
f(a)approx frac1315 a-1+O(a^2)
$$
Except the coefficient of $a^-1$, all the other orders are consistent.
divergent-series laurent-series divergence regularization
asked Jul 31 at 13:10
user142288
11
To find the Laurent series of function $f(a)$ at point $a=0$
$$
f(a)=int^1_0 fracd xx^2+a^2
$$
one can first do the integral
$$
f(a)=frac1aarctan(1/a)
$$
then expand $arctan(1/a)$ and obtain
$$
f(a)approx fracpi2a-1+O(a^2).
$$
However if one expands the integrand before integration,
one would get divergent integral.
My question is: is there any method helping us expand the function $f(a)$ before performing the integral?
The motivation arises from more complicated cases, when the functions can not be integrated out analytically.
==========================================
Updated: I think I find the solution to this integral. When I expand the function $f(a)$, I have to expand all the integral, not only the integrand
$$
frac1x^2+a^2approx frac1x^2+O(a^2)
$$
which means that the upper limit of integral has also to be changed, i.e.
$$
f(a)approx int^1_a d xleft(frac1x^2+O(a^2)right)
$$
this gives us
$$
f(a)approx frac1315 a-1+O(a^2)
$$
Except the coefficient of $a^-1$, all the other orders are consistent.
divergent-series laurent-series divergence regularization
asked Jul 31 at 13:10
user142288
11
edited 2 days ago
asked Jul 31 at 13:10
user142288
11
asked Jul 31 at 13:10
user142288
11
asked Jul 31 at 13:10
user142288
11
11
Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago
add a comment |Â
Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago
Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago
add a comment |Â
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Can you explain how exactly you expand the integrand?
– Batominovski
Jul 31 at 13:23
@Batominovski, $1/(x^2+a^2)approx 1/x^2-a^2/x^4+O(a^3)$, so every terms are divergent at lower limit.
– user142288
Jul 31 at 13:28
That's the expansion around $x=infty$ (valid in complex analysis). The expansion around $x=0$ is given by $$frac1a^2+x^2 = frac1a^2 frac11+x^2/a^2 = frac1a^2 left( 1 - fracx^2a^2 + O(x^4) right) = frac1a^2 - fracx^2a^4 + O(x^4)$$
– md2perpe
2 days ago