Mixing
Selecting functions expressed with complex analysis
I just read the identity, for $x>0$,
$$frac{1}{2ipi} int_{x-iinfty}^{x+iinfty} y^s frac{ds}{s} =
left{
begin{array}{cl}
1 & text{if } y > 1 \
0 & text{if } 0 < y < 1 \
end{array}
right.
$$
It do not understand what this integral is true, nor have I much clues on how to address such complex integrals. It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem? I do not understand how it helps here.
complex-analysis
asked Nov 20 '18 at 0:25
TheStudent
1886
add a comment |
I just read the identity, for $x>0$,
$$frac{1}{2ipi} int_{x-iinfty}^{x+iinfty} y^s frac{ds}{s} =
left{
begin{array}{cl}
1 & text{if } y > 1 \
0 & text{if } 0 < y < 1 \
end{array}
right.
$$
It do not understand what this integral is true, nor have I much clues on how to address such complex integrals. It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem? I do not understand how it helps here.
complex-analysis
asked Nov 20 '18 at 0:25
TheStudent
1886
" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
1
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14
add a comment |
I just read the identity, for $x>0$,
$$frac{1}{2ipi} int_{x-iinfty}^{x+iinfty} y^s frac{ds}{s} =
left{
begin{array}{cl}
1 & text{if } y > 1 \
0 & text{if } 0 < y < 1 \
end{array}
right.
$$
It do not understand what this integral is true, nor have I much clues on how to address such complex integrals. It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem? I do not understand how it helps here.
complex-analysis
asked Nov 20 '18 at 0:25
TheStudent
1886
I just read the identity, for $x>0$,
$$frac{1}{2ipi} int_{x-iinfty}^{x+iinfty} y^s frac{ds}{s} =
left{
begin{array}{cl}
1 & text{if } y > 1 \
0 & text{if } 0 < y < 1 \
end{array}
right.
$$
It do not understand what this integral is true, nor have I much clues on how to address such complex integrals. It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem? I do not understand how it helps here.
complex-analysis
complex-analysis
asked Nov 20 '18 at 0:25
TheStudent
1886
asked Nov 20 '18 at 0:25
TheStudent
1886
edited Nov 20 '18 at 8:57
asked Nov 20 '18 at 0:25
TheStudent
1886
asked Nov 20 '18 at 0:25
TheStudent
1886
asked Nov 20 '18 at 0:25
TheStudent
1886
1886
" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
1
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14
add a comment |
" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
1
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14
" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
1
1
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14
add a comment |
1 Answer
1
active
oldest
votes
Let $b=ln y$.
Then, your integral is
$$I(b)=lim_{Rtoinfty}int^{x+iR}_{x-iR}e^{bs}s^{-1}ds$$
When $b>0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, frac{pi}2le tle frac{3pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the left, forming a closed contour.
By residue theorem,
$$oint_{Gamma}e^{bs}s^{-1}ds=2pi ioperatorname*{Res}_{s=0}e^{bs}s^{-1}=2pi i$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{3pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{3pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{3pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{pi}2,frac{3pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=2pi i$ when $b>0$.
When $b<0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, -frac{pi}2le tle frac{pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the right, forming a closed contour.
Since no singularities are enclosed,
$$oint_{Gamma}e^{bs}s^{-1}ds=0$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{-pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{-pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{-pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{-pi}2,frac{pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=0$ when $b>0$.
Assembling everything we have,
$$int^{x+iinfty}_{x-iinfty}e^{bs}s^{-1}ds=
begin{cases}
2pi i, &b>0 \
0, &-infty<b<0 \
end{cases}
$$
Substituting in $b=ln y$ and dividing both sides by $2pi i$ one recovers
$$color{red}{
frac1{2pi i}int^{x+iinfty}_{x-iinfty}y^{s}frac{ds}s=
begin{cases}
1, &y>1 \
0, &0<y<1 \
end{cases}
}
$$
as expected.
answered Nov 20 '18 at 6:23
Szeto
6,4362926
add a comment |
Your Answer
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1 Answer
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1 Answer
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oldest
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oldest
votes
Let $b=ln y$.
Then, your integral is
$$I(b)=lim_{Rtoinfty}int^{x+iR}_{x-iR}e^{bs}s^{-1}ds$$
When $b>0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, frac{pi}2le tle frac{3pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the left, forming a closed contour.
By residue theorem,
$$oint_{Gamma}e^{bs}s^{-1}ds=2pi ioperatorname*{Res}_{s=0}e^{bs}s^{-1}=2pi i$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{3pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{3pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{3pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{pi}2,frac{3pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=2pi i$ when $b>0$.
When $b<0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, -frac{pi}2le tle frac{pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the right, forming a closed contour.
Since no singularities are enclosed,
$$oint_{Gamma}e^{bs}s^{-1}ds=0$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{-pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{-pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{-pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{-pi}2,frac{pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=0$ when $b>0$.
Assembling everything we have,
$$int^{x+iinfty}_{x-iinfty}e^{bs}s^{-1}ds=
begin{cases}
2pi i, &b>0 \
0, &-infty<b<0 \
end{cases}
$$
Substituting in $b=ln y$ and dividing both sides by $2pi i$ one recovers
$$color{red}{
frac1{2pi i}int^{x+iinfty}_{x-iinfty}y^{s}frac{ds}s=
begin{cases}
1, &y>1 \
0, &0<y<1 \
end{cases}
}
$$
as expected.
answered Nov 20 '18 at 6:23
Szeto
6,4362926
add a comment |
Let $b=ln y$.
Then, your integral is
$$I(b)=lim_{Rtoinfty}int^{x+iR}_{x-iR}e^{bs}s^{-1}ds$$
When $b>0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, frac{pi}2le tle frac{3pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the left, forming a closed contour.
By residue theorem,
$$oint_{Gamma}e^{bs}s^{-1}ds=2pi ioperatorname*{Res}_{s=0}e^{bs}s^{-1}=2pi i$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{3pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{3pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{3pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{pi}2,frac{3pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=2pi i$ when $b>0$.
When $b<0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, -frac{pi}2le tle frac{pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the right, forming a closed contour.
Since no singularities are enclosed,
$$oint_{Gamma}e^{bs}s^{-1}ds=0$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{-pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{-pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{-pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{-pi}2,frac{pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=0$ when $b>0$.
Assembling everything we have,
$$int^{x+iinfty}_{x-iinfty}e^{bs}s^{-1}ds=
begin{cases}
2pi i, &b>0 \
0, &-infty<b<0 \
end{cases}
$$
Substituting in $b=ln y$ and dividing both sides by $2pi i$ one recovers
$$color{red}{
frac1{2pi i}int^{x+iinfty}_{x-iinfty}y^{s}frac{ds}s=
begin{cases}
1, &y>1 \
0, &0<y<1 \
end{cases}
}
$$
as expected.
answered Nov 20 '18 at 6:23
Szeto
6,4362926
add a comment |
Let $b=ln y$.
Then, your integral is
$$I(b)=lim_{Rtoinfty}int^{x+iR}_{x-iR}e^{bs}s^{-1}ds$$
When $b>0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, frac{pi}2le tle frac{3pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the left, forming a closed contour.
By residue theorem,
$$oint_{Gamma}e^{bs}s^{-1}ds=2pi ioperatorname*{Res}_{s=0}e^{bs}s^{-1}=2pi i$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{3pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{3pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{3pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{pi}2,frac{3pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=2pi i$ when $b>0$.
When $b<0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, -frac{pi}2le tle frac{pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the right, forming a closed contour.
Since no singularities are enclosed,
$$oint_{Gamma}e^{bs}s^{-1}ds=0$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{-pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{-pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{-pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{-pi}2,frac{pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=0$ when $b>0$.
Assembling everything we have,
$$int^{x+iinfty}_{x-iinfty}e^{bs}s^{-1}ds=
begin{cases}
2pi i, &b>0 \
0, &-infty<b<0 \
end{cases}
$$
Substituting in $b=ln y$ and dividing both sides by $2pi i$ one recovers
$$color{red}{
frac1{2pi i}int^{x+iinfty}_{x-iinfty}y^{s}frac{ds}s=
begin{cases}
1, &y>1 \
0, &0<y<1 \
end{cases}
}
$$
as expected.
answered Nov 20 '18 at 6:23
Szeto
6,4362926
Let $b=ln y$.
Then, your integral is
$$I(b)=lim_{Rtoinfty}int^{x+iR}_{x-iR}e^{bs}s^{-1}ds$$
When $b>0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, frac{pi}2le tle frac{3pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the left, forming a closed contour.
By residue theorem,
$$oint_{Gamma}e^{bs}s^{-1}ds=2pi ioperatorname*{Res}_{s=0}e^{bs}s^{-1}=2pi i$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{3pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{3pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{3pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{pi}2,frac{3pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=2pi i$ when $b>0$.
When $b<0$:
Pick a contour $Gamma(R)=gamma_1cupgamma_2$, parametrized by
$$gamma_1:~~~~s(t)=x+it, -Rle tle R$$
$$gamma_2:~~~~s(t)=x+Re^{it}, -frac{pi}2le tle frac{pi}2$$
If you draw it out, it is the integration path of $I(b)$ plus a semicircle on the right, forming a closed contour.
Since no singularities are enclosed,
$$oint_{Gamma}e^{bs}s^{-1}ds=0$$
Also,
$$begin{align}
left|int_{gamma_2}e^{bs}s^{-1}dsright|
&=left|int^{-pi/2}_{pi/2}e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}dt right|\
&le int^{-pi/2}_{pi/2}left|e^{bx}~e^{bRexp(it)}frac{iRe^{it}}{x+Re^{it}}right|dt \
&=int^{-pi/2}_{pi/2}e^{bx}~e^{bRcos t}frac{R}{|x+Re^{it}|}dt \
&=pi e^{bx}~e^{bRcos c}frac{R}{|x+Re^{ic}|} qquad{text{where $cinleft[frac{-pi}2,frac{pi}2right]$}} \
&xrightarrow{Rtoinfty} 0
end{align}
$$
Thus we have that $I(b)=0$ when $b>0$.
Assembling everything we have,
$$int^{x+iinfty}_{x-iinfty}e^{bs}s^{-1}ds=
begin{cases}
2pi i, &b>0 \
0, &-infty<b<0 \
end{cases}
$$
Substituting in $b=ln y$ and dividing both sides by $2pi i$ one recovers
$$color{red}{
frac1{2pi i}int^{x+iinfty}_{x-iinfty}y^{s}frac{ds}s=
begin{cases}
1, &y>1 \
0, &0<y<1 \
end{cases}
}
$$
as expected.
answered Nov 20 '18 at 6:23
Szeto
6,4362926
edited Nov 21 '18 at 3:40
answered Nov 20 '18 at 6:23
Szeto
6,4362926
answered Nov 20 '18 at 6:23
Szeto
6,4362926
answered Nov 20 '18 at 6:23
Szeto
6,4362926
6,4362926
add a comment |
add a comment |
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" It is said that it is true by "residue calculus", does this mean that we should use the Cauchy residue theorem?" Yesn this means that. Is this your question?
– Did
Nov 20 '18 at 7:04
1
@Did Please re-read the question more carefully. There are hidden questions not marked with a question mark. Please do forgive this fairly new user.
– Szeto
Nov 20 '18 at 8:46
@Szeto Thanks for this (quite inadequate) admonestation. The site is for explicit questions, especially when the OP is a beginner, not for hidden ones.
– Did
Nov 20 '18 at 15:14