Find using Residue Theorem the following integral
$begingroup$
Find using Residue Theorem
$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
My try:
I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.
and $C_2$ is the line joining $-1$ to $1$.
I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.
Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Any help from someone here?
Thanks for reading my post
complex-analysis residue-calculus
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
$endgroup$
add a comment |
$begingroup$
Find using Residue Theorem
$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
My try:
I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.
and $C_2$ is the line joining $-1$ to $1$.
I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.
Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Any help from someone here?
Thanks for reading my post
complex-analysis residue-calculus
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
$endgroup$
$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02
$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07
$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14
add a comment |
$begingroup$
Find using Residue Theorem
$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
My try:
I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.
and $C_2$ is the line joining $-1$ to $1$.
I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.
Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Any help from someone here?
Thanks for reading my post
complex-analysis residue-calculus
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
$endgroup$
Find using Residue Theorem
$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
My try:
I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$.
and $C_2$ is the line joining $-1$ to $1$.
I now consider the function in $Bbb C$ to be
$$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$=$$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$
But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$.
Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$
Any help from someone here?
Thanks for reading my post
complex-analysis residue-calculus
complex-analysis residue-calculus
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
edited Dec 4 '18 at 17:06
Join_PhD
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
asked Dec 4 '18 at 16:57
Join_PhDJoin_PhD
3898
3898
$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02
$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07
$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14
add a comment |
$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02
$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07
$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14
$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02
$begingroup$
Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
$endgroup$
– DonAntonio
Dec 4 '18 at 17:02
$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07
$begingroup$
@DonAntonio;edited my question,can u have a look now
$endgroup$
– Join_PhD
Dec 4 '18 at 17:07
$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14
$begingroup$
Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
$endgroup$
– DonAntonio
Dec 4 '18 at 18:14
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.
Let's pick a contour similar to the one seen in this example.
We pick the branch cut on $[-1,1]$ on the real line such that
begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}
The contour consists of:
$C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,
$C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$
$C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$
$C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$
In the limit of $epsilon to 0$, we have
$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$
Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality
$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1-z| ge 2$
$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1+z| ge 2$
Finally, use residues to finish the rest. You may also need to find the residue at infinity.
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
$endgroup$
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
add a comment |
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1 Answer
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1 Answer
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$begingroup$
The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.
Let's pick a contour similar to the one seen in this example.
We pick the branch cut on $[-1,1]$ on the real line such that
begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}
The contour consists of:
$C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,
$C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$
$C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$
$C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$
In the limit of $epsilon to 0$, we have
$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$
Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality
$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1-z| ge 2$
$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1+z| ge 2$
Finally, use residues to finish the rest. You may also need to find the residue at infinity.
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
$endgroup$
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
add a comment |
$begingroup$
The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.
Let's pick a contour similar to the one seen in this example.
We pick the branch cut on $[-1,1]$ on the real line such that
begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}
The contour consists of:
$C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,
$C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$
$C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$
$C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$
In the limit of $epsilon to 0$, we have
$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$
Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality
$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1-z| ge 2$
$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1+z| ge 2$
Finally, use residues to finish the rest. You may also need to find the residue at infinity.
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
$endgroup$
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
add a comment |
$begingroup$
The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.
Let's pick a contour similar to the one seen in this example.
We pick the branch cut on $[-1,1]$ on the real line such that
begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}
The contour consists of:
$C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,
$C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$
$C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$
$C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$
In the limit of $epsilon to 0$, we have
$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$
Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality
$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1-z| ge 2$
$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1+z| ge 2$
Finally, use residues to finish the rest. You may also need to find the residue at infinity.
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
$endgroup$
The complex function has two branch points at $z=pm 1$ and a pole at $z=i$. Your contour contains all of these points, making it an inappropriate contour to use.
Let's pick a contour similar to the one seen in this example.
We pick the branch cut on $[-1,1]$ on the real line such that
begin{align}
1+z &= r_1e^{iphi_1}, & phi_1 in (-pi,pi] \
1-z &= r_2e^{iphi_2}, & phi_2 in (0,2pi)
end{align}
The contour consists of:
$C_1$: a line segment just above the branch cut going from $1+epsilon$ to $-1+epsilon$,
$C_2$: a left semicircle $z = -1 +epsilon e^{it}$, going counter-clockwise from $-1+epsilon$ to $-1-epsilon$
$C_3$: a line segment just below the branch cut, going from $-1-epsilon$ to $1-epsilon$
$C_4$: a right semicircle $z = 1 + epsilon e^{it}$, going counterclockwise from $1-epsilon$ to $1+epsilon$
In the limit of $epsilon to 0$, we have
$$ f(z)big|_{C_1} = frac{1}{(z^2+1)sqrt{|1+z|}e^{i0/2}sqrt{|1-z|}e^{i2pi/2}} = frac{-1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ f(z)big|_{C_3} = frac{1}{(z^2+1)sqrt{|1+z|}e^{-i0/2}sqrt{|1-z|}e^{i0/2}} = frac{1}{(z^2+1)sqrt{|1+z|}sqrt{|1-z|}} $$
$$ implies int_{C_1} f(z) dz + int_{C_3} f(z) dz = -int_1^{-1} f(x) dx + int_{-1}^1 f(x) dx = 2int_{-1}^1 f(x) dx $$
Next, we can prove the integral vanishes on the 2 semicircles. Using the ML inequality
$$ int_{C_2} f(z) dz le frac{L(C_2)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1-z| ge 2$
$$ int_{C_4} f(z) dz le frac{L(C_4)}{|1+z^2|sqrt{|1-z|}sqrt{|1+z|}} le frac{pi epsilon}{2sqrt{2}sqrt{epsilon}} to 0 $$
Since $|z| ge 1$ and $|1+z| ge 2$
Finally, use residues to finish the rest. You may also need to find the residue at infinity.
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
answered Dec 5 '18 at 10:07
DylanDylan
12.7k31026
12.7k31026
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
add a comment |
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
$begingroup$
Thanks for an answer,I will need some time and patience to understand it
$endgroup$
– Join_PhD
Dec 5 '18 at 10:09
add a comment |
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Either all the time there is lacking some parentheses in the denominator or else there is rather weird $;1;$ there...
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– DonAntonio
Dec 4 '18 at 17:02
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@DonAntonio;edited my question,can u have a look now
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– Join_PhD
Dec 4 '18 at 17:07
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Have you already chosen a branch for the square root function? There are two of them...But this isn't really that critical now: you cannot have a path from the point $;z=-1;$ to the point $;z=1;$ since on those points your function isn't defined. You can't also go through the point $;z=i;$ for the same reason.
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– DonAntonio
Dec 4 '18 at 18:14