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How to solve this complex integral?
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I need to solve the equation
$I = int_0^ufracdx(a + bx + iepsilon)^2$,
where $a$ and $b$ are real, and functions of $u$ ($u$ is another variable that I have to perform another integral after). I tried to use the Sokhotski–Plemelj theorem,
$lim_epsilonto 0int_-infty^+inftyfracf(x)dxx-x_0+iepsilon = Pint_-infty^+inftyfracf(x)dxx-x_0 + ipi delta(x - x_0)$
I don't know if I can directly use this theorem, once the integration limits in $I$ are different. If there is no problem, I have the Principal Value, $fracua(a-bu)$, but I am not sure about the argument of delta function. There is another way to proceed? Thanks in advance!
contour-integration complex-integration
asked Jul 26 at 13:50
Dyana Duarte
61
 |Â
show 1 more comment
up vote
1
down vote
favorite
I need to solve the equation
$I = int_0^ufracdx(a + bx + iepsilon)^2$,
where $a$ and $b$ are real, and functions of $u$ ($u$ is another variable that I have to perform another integral after). I tried to use the Sokhotski–Plemelj theorem,
$lim_epsilonto 0int_-infty^+inftyfracf(x)dxx-x_0+iepsilon = Pint_-infty^+inftyfracf(x)dxx-x_0 + ipi delta(x - x_0)$
I don't know if I can directly use this theorem, once the integration limits in $I$ are different. If there is no problem, I have the Principal Value, $fracua(a-bu)$, but I am not sure about the argument of delta function. There is another way to proceed? Thanks in advance!
contour-integration complex-integration
asked Jul 26 at 13:50
Dyana Duarte
61
Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
1
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32
 |Â
show 1 more comment
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to solve the equation
$I = int_0^ufracdx(a + bx + iepsilon)^2$,
where $a$ and $b$ are real, and functions of $u$ ($u$ is another variable that I have to perform another integral after). I tried to use the Sokhotski–Plemelj theorem,
$lim_epsilonto 0int_-infty^+inftyfracf(x)dxx-x_0+iepsilon = Pint_-infty^+inftyfracf(x)dxx-x_0 + ipi delta(x - x_0)$
I don't know if I can directly use this theorem, once the integration limits in $I$ are different. If there is no problem, I have the Principal Value, $fracua(a-bu)$, but I am not sure about the argument of delta function. There is another way to proceed? Thanks in advance!
contour-integration complex-integration
asked Jul 26 at 13:50
Dyana Duarte
61
I need to solve the equation
$I = int_0^ufracdx(a + bx + iepsilon)^2$,
where $a$ and $b$ are real, and functions of $u$ ($u$ is another variable that I have to perform another integral after). I tried to use the Sokhotski–Plemelj theorem,
$lim_epsilonto 0int_-infty^+inftyfracf(x)dxx-x_0+iepsilon = Pint_-infty^+inftyfracf(x)dxx-x_0 + ipi delta(x - x_0)$
I don't know if I can directly use this theorem, once the integration limits in $I$ are different. If there is no problem, I have the Principal Value, $fracua(a-bu)$, but I am not sure about the argument of delta function. There is another way to proceed? Thanks in advance!
contour-integration complex-integration
asked Jul 26 at 13:50
Dyana Duarte
61
asked Jul 26 at 13:50
Dyana Duarte
61
asked Jul 26 at 13:50
Dyana Duarte
61
asked Jul 26 at 13:50
Dyana Duarte
61
61
Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
1
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32
 |Â
show 1 more comment
Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
1
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32
Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
1
1
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32
 |Â
show 1 more comment
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Mathematica 11.3 thinks it should be $$ I=fracu(a+i varepsilon)(a+bu+i varepsilon) $$
– Benedict W. J. Irwin
Jul 26 at 14:44
Yes, I already tried on Mathematica... But this $i epsilon$ is used to avoid the singularity, and the program treats it as a simple variable. That's why I am not sure if it is correct.
– Dyana Duarte
Jul 26 at 15:47
What singularity are you trying to avoid? There are none in your integral range.
– Von Neumann
Jul 26 at 16:11
In this case there are none singularity because $i epsilon$ avoids it... My question is how to solve this integral, to obtain an expression that is a function of $u$!
– Dyana Duarte
Jul 26 at 16:59
1
The epsilon is a simple variable. You can evaluate the integral in closed form and then take the limit: $$lim_epsilon downarrow 0 int_0^u frac dx (a + b x + i epsilon)^2 = lim_epsilon downarrow 0 frac u (a + i epsilon) (a + b u + i epsilon) = frac u a(a + b u).$$ The limit exists as long as $x=−a/b$ is not one of the endpoints of $[0, u]$. Note that your Sokhotski–Plemelj is not quite correct: by linearity, the last term cannot be independent of $f$.
– Maxim
Jul 26 at 20:32