Verification for $int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz$
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked yesterday
aleden
1,775411
|
show 2 more comments
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked yesterday
aleden
1,775411
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday
|
show 2 more comments
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
asked yesterday
aleden
1,775411
I decided to tackle this integral by integrating over a semicircular contour in the upper-half plane and the real line, where the path is denoted as $Gamma$.
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=(int_{C_R}+int_{-R}^R) frac{e^{iaz}}{z^{2n}+1}dz$$
The integral over $C_R$ should go to $0$ as $Rtoinfty$ by Jordan's Lemma. Thus
$$int_{Gamma} frac{e^{iaz}}{z^{2n}+1}dz=int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=2pi isum Res$$ by Residue Theorem. The function has a pole at $z_0=e^{frac{pi i}{2n}}$ and calculating the Residue at $z_0$ I get
$$lim_{zto z_0} e^{iaz} frac{z-z_0}{z^{2n}+1}=-e^{iaz_0}frac{z_0}{2n}$$
So $$int_{-infty}^infty frac{e^{iaz}}{z^{2n}+1}dz=-frac{pi i}{n}z_0e^{iaz_0}$$
After equating the real and imaginary terms on both sides I get that $$int_{-infty}^infty frac{cos(az)}{z^{2n}+1}dz=frac{pi}{n}e^{-asin(frac{pi}{2n})}(cos(frac{pi}{2n})sin(acos(frac{pi}{2n}))+sin(frac{pi}{2n})cos(acos(frac{pi}{2n})))$$ I found that this answer agrees with the Wolfram for $n=1$ but appears to be off by a factor of $2$ for $n=2$. Could anyone explain if/where I made a mistake and what the final answer should be?
proof-verification improper-integrals contour-integration
proof-verification improper-integrals contour-integration
asked yesterday
aleden
1,775411
asked yesterday
aleden
1,775411
asked yesterday
aleden
1,775411
asked yesterday
aleden
1,775411
asked yesterday
aleden
1,775411
1,775411
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday
|
show 2 more comments
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday
3
3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
1
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
1
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday
|
show 2 more comments
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3
There are $n$ poles in the upper half-plane. You appear only to be considering one of them.
– Lord Shark the Unknown
yesterday
@LordSharktheUnknown What would the other poles be?
– aleden
yesterday
@aleden: all the roots of $z^{2n}+1$ in the upper half-plane.
– Jack D'Aurizio
yesterday
1
@aleden: $z^{2n}+1$ is a divisor of $z^{4n}-1$.
– Jack D'Aurizio
yesterday
1
don't believe too much on wolfram alpha, neither mathematica
– Masacroso
yesterday