Mixing
Contour integration - complex analysis
I am trying to solve the following integral using a contour (large semi-circle connected to smaller semi-circle in the upper-half plane):
$$int_0^{infty} frac{log^4(x)}{1+x^2} dx.$$
I have split the contour into 4 parts - the large semi-circle, the small semi-circle, the part on the negative real axis and the part on the positive real axis.
The integral of the function over the contour is $2pi i sum Res(f)$, which is $pi^5$. The function has poles: $i$ and $-i$, each of order $1$, but $i$ is the only pole contained in the contour.
The integral over the large semi-circle is $0$ as the large radius approaches infinity and the integral over the small semi-circle is $0$ as the small radius approaches $0$.
I take the real part of both sides and the following is left:
$$
pi^5 = 2int_0^{infty} frac{log^4(x)}{1+x^2} dx
+ int_{-infty}^0 frac{-6pi^2log^2(x) + pi^4}{1+x^2} dx
$$
My final answer is $5pi^5/8$, but the correct answer is $5pi^5/32$.
Any suggestions? Thank you!
integration complex-analysis contour-integration complex-integration
edited Nov 21 '18 at 5:00
gt6989b
33k22452
asked Nov 21 '18 at 4:27
math_b
84
add a comment |
I am trying to solve the following integral using a contour (large semi-circle connected to smaller semi-circle in the upper-half plane):
$$int_0^{infty} frac{log^4(x)}{1+x^2} dx.$$
I have split the contour into 4 parts - the large semi-circle, the small semi-circle, the part on the negative real axis and the part on the positive real axis.
The integral of the function over the contour is $2pi i sum Res(f)$, which is $pi^5$. The function has poles: $i$ and $-i$, each of order $1$, but $i$ is the only pole contained in the contour.
The integral over the large semi-circle is $0$ as the large radius approaches infinity and the integral over the small semi-circle is $0$ as the small radius approaches $0$.
I take the real part of both sides and the following is left:
$$
pi^5 = 2int_0^{infty} frac{log^4(x)}{1+x^2} dx
+ int_{-infty}^0 frac{-6pi^2log^2(x) + pi^4}{1+x^2} dx
$$
My final answer is $5pi^5/8$, but the correct answer is $5pi^5/32$.
Any suggestions? Thank you!
integration complex-analysis contour-integration complex-integration
edited Nov 21 '18 at 5:00
gt6989b
33k22452
asked Nov 21 '18 at 4:27
math_b
84
1
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50
add a comment |
I am trying to solve the following integral using a contour (large semi-circle connected to smaller semi-circle in the upper-half plane):
$$int_0^{infty} frac{log^4(x)}{1+x^2} dx.$$
I have split the contour into 4 parts - the large semi-circle, the small semi-circle, the part on the negative real axis and the part on the positive real axis.
The integral of the function over the contour is $2pi i sum Res(f)$, which is $pi^5$. The function has poles: $i$ and $-i$, each of order $1$, but $i$ is the only pole contained in the contour.
The integral over the large semi-circle is $0$ as the large radius approaches infinity and the integral over the small semi-circle is $0$ as the small radius approaches $0$.
I take the real part of both sides and the following is left:
$$
pi^5 = 2int_0^{infty} frac{log^4(x)}{1+x^2} dx
+ int_{-infty}^0 frac{-6pi^2log^2(x) + pi^4}{1+x^2} dx
$$
My final answer is $5pi^5/8$, but the correct answer is $5pi^5/32$.
Any suggestions? Thank you!
integration complex-analysis contour-integration complex-integration
edited Nov 21 '18 at 5:00
gt6989b
33k22452
asked Nov 21 '18 at 4:27
math_b
84
I am trying to solve the following integral using a contour (large semi-circle connected to smaller semi-circle in the upper-half plane):
$$int_0^{infty} frac{log^4(x)}{1+x^2} dx.$$
I have split the contour into 4 parts - the large semi-circle, the small semi-circle, the part on the negative real axis and the part on the positive real axis.
The integral of the function over the contour is $2pi i sum Res(f)$, which is $pi^5$. The function has poles: $i$ and $-i$, each of order $1$, but $i$ is the only pole contained in the contour.
The integral over the large semi-circle is $0$ as the large radius approaches infinity and the integral over the small semi-circle is $0$ as the small radius approaches $0$.
I take the real part of both sides and the following is left:
$$
pi^5 = 2int_0^{infty} frac{log^4(x)}{1+x^2} dx
+ int_{-infty}^0 frac{-6pi^2log^2(x) + pi^4}{1+x^2} dx
$$
My final answer is $5pi^5/8$, but the correct answer is $5pi^5/32$.
Any suggestions? Thank you!
integration complex-analysis contour-integration complex-integration
integration complex-analysis contour-integration complex-integration
edited Nov 21 '18 at 5:00
gt6989b
33k22452
asked Nov 21 '18 at 4:27
math_b
84
edited Nov 21 '18 at 5:00
gt6989b
33k22452
asked Nov 21 '18 at 4:27
math_b
84
edited Nov 21 '18 at 5:00
gt6989b
33k22452
edited Nov 21 '18 at 5:00
gt6989b
33k22452
edited Nov 21 '18 at 5:00
gt6989b
33k22452
33k22452
asked Nov 21 '18 at 4:27
math_b
84
asked Nov 21 '18 at 4:27
math_b
84
asked Nov 21 '18 at 4:27
math_b
84
84
1
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50
add a comment |
1
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50
1
1
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50
add a comment |
1 Answer
1
active
oldest
votes
Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=int^infty_0frac{log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=int^infty_{0}frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^aequiv exp(a(ln|z|+iarg z))$ where $arg zin[0,2pi)$.
Firstly, by residue theorem,
$$oint_C f(z)dz=2pi ibigg(operatorname*{Res}_{z=i}f(z)+operatorname*{Res}_{z=-i}f(z)bigg)$$
We have
$$operatorname*{Res}_{z=i}f(z)=frac{exp(a(ln|i|+iarg i))}{i+i}=frac{e^{pi ia/2}}{2i}$$
$$operatorname*{Res}_{z=-i}f(z)=frac{exp(a(ln|-i|+iarg -i))}{-i-i}=-frac{e^{3pi ia/2}}{2i}$$
Thus,
$$oint_C f(z)dz=pi(e^{pi ia/2}-e^{3pi ia/2})$$
On the other hand,
$$oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=lim_{Rtoinfty}int^{2pi}_0 f(Re^{it})iRe^{it}dt
=lim_{Rtoinfty}2pi f(Re^{ic})iRe^{ic}=0 qquad{cin[0,2pi]}$$
$$K_2=lim_{rto0^+}int_{2pi}^0 f(re^{it})ire^{it}dt
=lim_{rto0^+}2pi f(re^{ic})ire^{ic}=0 qquad{cin[0,2pi]}$$
$$K_3=int^infty_0 f(te^{i0})dt=int^infty_0frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=int_infty^0 f(te^{i2pi})dt=-int^infty_0frac{e^{2pi ia}t^a}{t^2+1}dt=-e^{2pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2pi ia}I=pi(e^{pi ia/2}-e^{3pi ia/2})$$
$$implies I=pifrac{e^{pi ia/2}-e^{3pi ia/2}}{1-e^{2pi ia}}
=pifrac{e^{-pi ia/2}-e^{pi ia/2}}{e^{-pi i a}-e^{pi ia}}
=pifrac{sin(pi a/2)}{sin(pi a)}
=frac{pi}2secleft(frac{pi a}2right)
$$
Let $T=tan(pi a/2)$, $S=sec(pi a/2)$.
$$I^{(4)}(a)=frac{pi}2frac{pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=frac{pi}2frac{pi^4(0+0+5cdot 1)}{16}=color{red}{frac{5pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
answered Nov 22 '18 at 2:50
Szeto
6,4362926
add a comment |
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1 Answer
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Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=int^infty_0frac{log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=int^infty_{0}frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^aequiv exp(a(ln|z|+iarg z))$ where $arg zin[0,2pi)$.
Firstly, by residue theorem,
$$oint_C f(z)dz=2pi ibigg(operatorname*{Res}_{z=i}f(z)+operatorname*{Res}_{z=-i}f(z)bigg)$$
We have
$$operatorname*{Res}_{z=i}f(z)=frac{exp(a(ln|i|+iarg i))}{i+i}=frac{e^{pi ia/2}}{2i}$$
$$operatorname*{Res}_{z=-i}f(z)=frac{exp(a(ln|-i|+iarg -i))}{-i-i}=-frac{e^{3pi ia/2}}{2i}$$
Thus,
$$oint_C f(z)dz=pi(e^{pi ia/2}-e^{3pi ia/2})$$
On the other hand,
$$oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=lim_{Rtoinfty}int^{2pi}_0 f(Re^{it})iRe^{it}dt
=lim_{Rtoinfty}2pi f(Re^{ic})iRe^{ic}=0 qquad{cin[0,2pi]}$$
$$K_2=lim_{rto0^+}int_{2pi}^0 f(re^{it})ire^{it}dt
=lim_{rto0^+}2pi f(re^{ic})ire^{ic}=0 qquad{cin[0,2pi]}$$
$$K_3=int^infty_0 f(te^{i0})dt=int^infty_0frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=int_infty^0 f(te^{i2pi})dt=-int^infty_0frac{e^{2pi ia}t^a}{t^2+1}dt=-e^{2pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2pi ia}I=pi(e^{pi ia/2}-e^{3pi ia/2})$$
$$implies I=pifrac{e^{pi ia/2}-e^{3pi ia/2}}{1-e^{2pi ia}}
=pifrac{e^{-pi ia/2}-e^{pi ia/2}}{e^{-pi i a}-e^{pi ia}}
=pifrac{sin(pi a/2)}{sin(pi a)}
=frac{pi}2secleft(frac{pi a}2right)
$$
Let $T=tan(pi a/2)$, $S=sec(pi a/2)$.
$$I^{(4)}(a)=frac{pi}2frac{pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=frac{pi}2frac{pi^4(0+0+5cdot 1)}{16}=color{red}{frac{5pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
answered Nov 22 '18 at 2:50
Szeto
6,4362926
add a comment |
Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=int^infty_0frac{log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=int^infty_{0}frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^aequiv exp(a(ln|z|+iarg z))$ where $arg zin[0,2pi)$.
Firstly, by residue theorem,
$$oint_C f(z)dz=2pi ibigg(operatorname*{Res}_{z=i}f(z)+operatorname*{Res}_{z=-i}f(z)bigg)$$
We have
$$operatorname*{Res}_{z=i}f(z)=frac{exp(a(ln|i|+iarg i))}{i+i}=frac{e^{pi ia/2}}{2i}$$
$$operatorname*{Res}_{z=-i}f(z)=frac{exp(a(ln|-i|+iarg -i))}{-i-i}=-frac{e^{3pi ia/2}}{2i}$$
Thus,
$$oint_C f(z)dz=pi(e^{pi ia/2}-e^{3pi ia/2})$$
On the other hand,
$$oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=lim_{Rtoinfty}int^{2pi}_0 f(Re^{it})iRe^{it}dt
=lim_{Rtoinfty}2pi f(Re^{ic})iRe^{ic}=0 qquad{cin[0,2pi]}$$
$$K_2=lim_{rto0^+}int_{2pi}^0 f(re^{it})ire^{it}dt
=lim_{rto0^+}2pi f(re^{ic})ire^{ic}=0 qquad{cin[0,2pi]}$$
$$K_3=int^infty_0 f(te^{i0})dt=int^infty_0frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=int_infty^0 f(te^{i2pi})dt=-int^infty_0frac{e^{2pi ia}t^a}{t^2+1}dt=-e^{2pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2pi ia}I=pi(e^{pi ia/2}-e^{3pi ia/2})$$
$$implies I=pifrac{e^{pi ia/2}-e^{3pi ia/2}}{1-e^{2pi ia}}
=pifrac{e^{-pi ia/2}-e^{pi ia/2}}{e^{-pi i a}-e^{pi ia}}
=pifrac{sin(pi a/2)}{sin(pi a)}
=frac{pi}2secleft(frac{pi a}2right)
$$
Let $T=tan(pi a/2)$, $S=sec(pi a/2)$.
$$I^{(4)}(a)=frac{pi}2frac{pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=frac{pi}2frac{pi^4(0+0+5cdot 1)}{16}=color{red}{frac{5pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
answered Nov 22 '18 at 2:50
Szeto
6,4362926
add a comment |
Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=int^infty_0frac{log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=int^infty_{0}frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^aequiv exp(a(ln|z|+iarg z))$ where $arg zin[0,2pi)$.
Firstly, by residue theorem,
$$oint_C f(z)dz=2pi ibigg(operatorname*{Res}_{z=i}f(z)+operatorname*{Res}_{z=-i}f(z)bigg)$$
We have
$$operatorname*{Res}_{z=i}f(z)=frac{exp(a(ln|i|+iarg i))}{i+i}=frac{e^{pi ia/2}}{2i}$$
$$operatorname*{Res}_{z=-i}f(z)=frac{exp(a(ln|-i|+iarg -i))}{-i-i}=-frac{e^{3pi ia/2}}{2i}$$
Thus,
$$oint_C f(z)dz=pi(e^{pi ia/2}-e^{3pi ia/2})$$
On the other hand,
$$oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=lim_{Rtoinfty}int^{2pi}_0 f(Re^{it})iRe^{it}dt
=lim_{Rtoinfty}2pi f(Re^{ic})iRe^{ic}=0 qquad{cin[0,2pi]}$$
$$K_2=lim_{rto0^+}int_{2pi}^0 f(re^{it})ire^{it}dt
=lim_{rto0^+}2pi f(re^{ic})ire^{ic}=0 qquad{cin[0,2pi]}$$
$$K_3=int^infty_0 f(te^{i0})dt=int^infty_0frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=int_infty^0 f(te^{i2pi})dt=-int^infty_0frac{e^{2pi ia}t^a}{t^2+1}dt=-e^{2pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2pi ia}I=pi(e^{pi ia/2}-e^{3pi ia/2})$$
$$implies I=pifrac{e^{pi ia/2}-e^{3pi ia/2}}{1-e^{2pi ia}}
=pifrac{e^{-pi ia/2}-e^{pi ia/2}}{e^{-pi i a}-e^{pi ia}}
=pifrac{sin(pi a/2)}{sin(pi a)}
=frac{pi}2secleft(frac{pi a}2right)
$$
Let $T=tan(pi a/2)$, $S=sec(pi a/2)$.
$$I^{(4)}(a)=frac{pi}2frac{pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=frac{pi}2frac{pi^4(0+0+5cdot 1)}{16}=color{red}{frac{5pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
answered Nov 22 '18 at 2:50
Szeto
6,4362926
Feynman isn't necessarily antangonistic to contour integration. They sometimes work together well.
$$J=int^infty_0frac{log^4(x)}{1+x^2}dx=I^{(4)}(0)$$
where
$$I(a)=int^infty_{0}frac{x^{a}}{1+x^2}dx$$ with $-1<a<1$.
Take $C$ as a keyhole contour, centered at the origin, avoiding the positive real axis.
Let $f(z)=z^a(1+z^2)^{-1}$ with branch cut on the positive real axis, implying $z^aequiv exp(a(ln|z|+iarg z))$ where $arg zin[0,2pi)$.
Firstly, by residue theorem,
$$oint_C f(z)dz=2pi ibigg(operatorname*{Res}_{z=i}f(z)+operatorname*{Res}_{z=-i}f(z)bigg)$$
We have
$$operatorname*{Res}_{z=i}f(z)=frac{exp(a(ln|i|+iarg i))}{i+i}=frac{e^{pi ia/2}}{2i}$$
$$operatorname*{Res}_{z=-i}f(z)=frac{exp(a(ln|-i|+iarg -i))}{-i-i}=-frac{e^{3pi ia/2}}{2i}$$
Thus,
$$oint_C f(z)dz=pi(e^{pi ia/2}-e^{3pi ia/2})$$
On the other hand,
$$oint f(z)dz=K_1+K_2+K_3+K_4$$
where
$$
K_1=lim_{Rtoinfty}int^{2pi}_0 f(Re^{it})iRe^{it}dt
=lim_{Rtoinfty}2pi f(Re^{ic})iRe^{ic}=0 qquad{cin[0,2pi]}$$
$$K_2=lim_{rto0^+}int_{2pi}^0 f(re^{it})ire^{it}dt
=lim_{rto0^+}2pi f(re^{ic})ire^{ic}=0 qquad{cin[0,2pi]}$$
$$K_3=int^infty_0 f(te^{i0})dt=int^infty_0frac{e^{i0}t^a}{t^2+1}dt=I$$
$$K_4=int_infty^0 f(te^{i2pi})dt=-int^infty_0frac{e^{2pi ia}t^a}{t^2+1}dt=-e^{2pi ia}I$$
For $K_1,K_2$, please respectively note the asymptotics $f(z)sim z^{a}$ for small $|z|$ and $f(z)=O(z^{a-2})$ for large $|z|$.
Therefore,
$$I-e^{2pi ia}I=pi(e^{pi ia/2}-e^{3pi ia/2})$$
$$implies I=pifrac{e^{pi ia/2}-e^{3pi ia/2}}{1-e^{2pi ia}}
=pifrac{e^{-pi ia/2}-e^{pi ia/2}}{e^{-pi i a}-e^{pi ia}}
=pifrac{sin(pi a/2)}{sin(pi a)}
=frac{pi}2secleft(frac{pi a}2right)
$$
Let $T=tan(pi a/2)$, $S=sec(pi a/2)$.
$$I^{(4)}(a)=frac{pi}2frac{pi^4(T^4+18S^2T^2+5S^4)}{16}$$
Hence,
$$J=I^{(4)}(0)=frac{pi}2frac{pi^4(0+0+5cdot 1)}{16}=color{red}{frac{5pi^5}{32}}$$
The tedious differentiation is done by calculator. :)
answered Nov 22 '18 at 2:50
Szeto
6,4362926
edited Nov 22 '18 at 3:13
answered Nov 22 '18 at 2:50
Szeto
6,4362926
answered Nov 22 '18 at 2:50
Szeto
6,4362926
answered Nov 22 '18 at 2:50
Szeto
6,4362926
6,4362926
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1
In you argument how is $log, x$ defined for $x<0$?
– Kavi Rama Murthy
Nov 21 '18 at 4:56
for $ log(z)$, I have chosen the branch $frac{-pi}{2}$ $leq$ arg(z) $leq$ $frac{3pi}{2}$.
– math_b
Nov 21 '18 at 13:10
On the contour, when x $lt$ 0, it is just the line on the negative real axis (from $- R$ to $- r$), where $R$ is the radius of the larger semi-circle and $r$ is the radius of smaller semi-circle.
– math_b
Nov 21 '18 at 13:12
The following MSE link presents a procedure for higher powers of the logarithm.
– Marko Riedel
Nov 21 '18 at 13:50