Computing $lim_{epsilon rightarrow 0} int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}$
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I have to compute an integral similiar to a Kramers-Kronig, namely
$$ lim_{epsilon rightarrow 0} int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
$$
I tried some ways to calculate it, but ended up with various results, since I'm not sure when to properly take the limit.
I just show the straight forwards way i came up with. It should show most of my problems.
$$ int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
= -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma frac{a-gamma+i epsilon}{gamma^2} = -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma left(frac{a+i epsilon}{gamma^2} - frac{1}{gamma} right) = (a+i epsilon)left(frac{1}{a-c+i epsilon} - frac{1}{a-b+i epsilon}right) + logleft(frac{a-c+i epsilon}{a-b+i epsilon}right)
$$
Taking the limit $epsilon rightarrow 0$ that should get:
$$
frac{a(c-b)}{(a-c)(a-b)}+logleft(left|frac{a-c}{a-b}right|right) + i pi Thetaleft(-text{sign}left(frac{a-c}{a-b}right)right)
$$
Here i pulled out the minus out of the logarithm if a is between b and c.
integration definite-integrals
edited Jan 2 at 21:52
El borito
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
$endgroup$
add a comment |
$begingroup$
I have to compute an integral similiar to a Kramers-Kronig, namely
$$ lim_{epsilon rightarrow 0} int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
$$
I tried some ways to calculate it, but ended up with various results, since I'm not sure when to properly take the limit.
I just show the straight forwards way i came up with. It should show most of my problems.
$$ int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
= -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma frac{a-gamma+i epsilon}{gamma^2} = -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma left(frac{a+i epsilon}{gamma^2} - frac{1}{gamma} right) = (a+i epsilon)left(frac{1}{a-c+i epsilon} - frac{1}{a-b+i epsilon}right) + logleft(frac{a-c+i epsilon}{a-b+i epsilon}right)
$$
Taking the limit $epsilon rightarrow 0$ that should get:
$$
frac{a(c-b)}{(a-c)(a-b)}+logleft(left|frac{a-c}{a-b}right|right) + i pi Thetaleft(-text{sign}left(frac{a-c}{a-b}right)right)
$$
Here i pulled out the minus out of the logarithm if a is between b and c.
integration definite-integrals
edited Jan 2 at 21:52
El borito
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
$endgroup$
$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22
add a comment |
$begingroup$
I have to compute an integral similiar to a Kramers-Kronig, namely
$$ lim_{epsilon rightarrow 0} int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
$$
I tried some ways to calculate it, but ended up with various results, since I'm not sure when to properly take the limit.
I just show the straight forwards way i came up with. It should show most of my problems.
$$ int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
= -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma frac{a-gamma+i epsilon}{gamma^2} = -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma left(frac{a+i epsilon}{gamma^2} - frac{1}{gamma} right) = (a+i epsilon)left(frac{1}{a-c+i epsilon} - frac{1}{a-b+i epsilon}right) + logleft(frac{a-c+i epsilon}{a-b+i epsilon}right)
$$
Taking the limit $epsilon rightarrow 0$ that should get:
$$
frac{a(c-b)}{(a-c)(a-b)}+logleft(left|frac{a-c}{a-b}right|right) + i pi Thetaleft(-text{sign}left(frac{a-c}{a-b}right)right)
$$
Here i pulled out the minus out of the logarithm if a is between b and c.
integration definite-integrals
edited Jan 2 at 21:52
El borito
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
$endgroup$
I have to compute an integral similiar to a Kramers-Kronig, namely
$$ lim_{epsilon rightarrow 0} int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
$$
I tried some ways to calculate it, but ended up with various results, since I'm not sure when to properly take the limit.
I just show the straight forwards way i came up with. It should show most of my problems.
$$ int_{b}^{c} dx frac{x}{(a-x+i epsilon)^2}
= -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma frac{a-gamma+i epsilon}{gamma^2} = -int_{a-b+i epsilon}^{a-c+i epsilon} dgamma left(frac{a+i epsilon}{gamma^2} - frac{1}{gamma} right) = (a+i epsilon)left(frac{1}{a-c+i epsilon} - frac{1}{a-b+i epsilon}right) + logleft(frac{a-c+i epsilon}{a-b+i epsilon}right)
$$
Taking the limit $epsilon rightarrow 0$ that should get:
$$
frac{a(c-b)}{(a-c)(a-b)}+logleft(left|frac{a-c}{a-b}right|right) + i pi Thetaleft(-text{sign}left(frac{a-c}{a-b}right)right)
$$
Here i pulled out the minus out of the logarithm if a is between b and c.
integration definite-integrals
integration definite-integrals
edited Jan 2 at 21:52
El borito
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
edited Jan 2 at 21:52
El borito
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
edited Jan 2 at 21:52
El borito
664216
edited Jan 2 at 21:52
El borito
664216
edited Jan 2 at 21:52
El borito
664216
664216
asked Jan 2 at 20:36
M_kajM_kaj
112
asked Jan 2 at 20:36
M_kajM_kaj
112
asked Jan 2 at 20:36
M_kajM_kaj
112
112
$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22
add a comment |
$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22
$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22
$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22
add a comment |
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$begingroup$
Your result is correct if $a$ is not between $b$ and $c$. If $a$ is between $b$ and $c$, then the result is correct either if $b < c$ and $epsilon$ approaches zero from above or if $b > c$ and $epsilon$ approaches zero from below. Then the imaginary part of $(a - c + i epsilon)/(a - b + i epsilon)$ approaches zero from above. The real part approaches a negative value if $a$ is between $b$ and $c$. The conversion from $ln gamma_2 - ln gamma_1$ to $ln(gamma_2/gamma_1)$ is justified because $[gamma_1, gamma_2]$ doesn't cross the branch cut of the logarithm.
$endgroup$
– Maxim
Jan 3 at 2:22