Mixing
Solutions check. Find $int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt$
Let $x > 0$. Find $$lim_{b to infty} int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt$$
I solved this problem but I am not quite confident about my steps. So please, could you check my solutions and tell me if there is something wrong??. Thanks in advanced!
First approuch:
Let's just focus on the first member of the integral:
$$int_{-b}^{b} frac{1}{t + ix} dt =
int_{-b}^{b} frac{1}{t + ix} Big( frac{t - ix}{t - ix} Big) dt =
int_{-b}^{b} frac{t - ix}{t^2 + x^2} dt$$
where the integrand can be again separated:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt - int_{-b}^{b} frac{ix}{t^2 + x^2}dt$$
Then, since the integral is about a real number $t$, $i$ can be treated as a constant. Now, I notice that the second integrand is an even function so the integral from $-b$ to $b$ is $0$. So we have:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt =
frac{1}{2}lnBig(t^2 + x^2 Big) Bigrvert_{-b}^{b} =
frac{1}{2}lnBig(frac{b^2 + x^2}{b^2 + x^2} Big) = 0$$
Following almost the same steps one can also show that $$int_{-b}^{b} frac{1}{t - ix} dt= 0$$
So the question reminds as $$lim_{b to infty} (0 + 0) = 0$$
Is this correct?? I mean:
Can $i$ be really treated as a constat in an integral which is about a real number??
Why would the question ask for about a limit int the infinite if the integral was destined to be $0$??
Second approuch
If $i$ can be really treated as a constant then the integral of the function $frac{1}{t + ix}$ would be $ln(t + ix)$, since $ix$ would be also a constant, so we have:
$$int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt=
ln(t + ix)Bigrvert_{-b}^{b} - ln(t - ix)Bigrvert_{-b}^{b}$$
Since $ix$ is just a constant then $ln$ fullfills the same properties as the well-known real valued function $ln$, so:
$$ = lnBig(frac{b + ix}{-b + ix} Big) - lnBig(frac{b - ix}{-b - ix} Big) =
lnBig(frac{frac{b + ix}{-b + ix}}{frac{b - ix}{-b - ix}} Big) = ln(1) = 0$$
Soo the answer would be $lim_{b to infty} 0 = 0$. I am really confused with theese steps since $ln$ function was evaluated in $b + ix$ and although I know that there is a counter-part complex version of the real function $ln$, I had accepted $ix$ as a costant, so there wouldn't be nothing wrong, right??
Third approuch
If $i$ can be really treated as a constant, then:
$$int_{-b}^{b} Big( frac{1}{t+ ix} - frac{1}{t - ix} Big) dt =
int_{-b}^{b} Big( frac{-2ix}{t^2 + x^2} Big) dt$$
because the integrand is an even function, that integral is $0$. So we have again $lim_{b to infty} 0 = 0$
Any comment about what I wrote would be really appreciated!
complex-analysis proof-verification contour-integration
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
add a comment |
Let $x > 0$. Find $$lim_{b to infty} int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt$$
I solved this problem but I am not quite confident about my steps. So please, could you check my solutions and tell me if there is something wrong??. Thanks in advanced!
First approuch:
Let's just focus on the first member of the integral:
$$int_{-b}^{b} frac{1}{t + ix} dt =
int_{-b}^{b} frac{1}{t + ix} Big( frac{t - ix}{t - ix} Big) dt =
int_{-b}^{b} frac{t - ix}{t^2 + x^2} dt$$
where the integrand can be again separated:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt - int_{-b}^{b} frac{ix}{t^2 + x^2}dt$$
Then, since the integral is about a real number $t$, $i$ can be treated as a constant. Now, I notice that the second integrand is an even function so the integral from $-b$ to $b$ is $0$. So we have:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt =
frac{1}{2}lnBig(t^2 + x^2 Big) Bigrvert_{-b}^{b} =
frac{1}{2}lnBig(frac{b^2 + x^2}{b^2 + x^2} Big) = 0$$
Following almost the same steps one can also show that $$int_{-b}^{b} frac{1}{t - ix} dt= 0$$
So the question reminds as $$lim_{b to infty} (0 + 0) = 0$$
Is this correct?? I mean:
Can $i$ be really treated as a constat in an integral which is about a real number??
Why would the question ask for about a limit int the infinite if the integral was destined to be $0$??
Second approuch
If $i$ can be really treated as a constant then the integral of the function $frac{1}{t + ix}$ would be $ln(t + ix)$, since $ix$ would be also a constant, so we have:
$$int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt=
ln(t + ix)Bigrvert_{-b}^{b} - ln(t - ix)Bigrvert_{-b}^{b}$$
Since $ix$ is just a constant then $ln$ fullfills the same properties as the well-known real valued function $ln$, so:
$$ = lnBig(frac{b + ix}{-b + ix} Big) - lnBig(frac{b - ix}{-b - ix} Big) =
lnBig(frac{frac{b + ix}{-b + ix}}{frac{b - ix}{-b - ix}} Big) = ln(1) = 0$$
Soo the answer would be $lim_{b to infty} 0 = 0$. I am really confused with theese steps since $ln$ function was evaluated in $b + ix$ and although I know that there is a counter-part complex version of the real function $ln$, I had accepted $ix$ as a costant, so there wouldn't be nothing wrong, right??
Third approuch
If $i$ can be really treated as a constant, then:
$$int_{-b}^{b} Big( frac{1}{t+ ix} - frac{1}{t - ix} Big) dt =
int_{-b}^{b} Big( frac{-2ix}{t^2 + x^2} Big) dt$$
because the integrand is an even function, that integral is $0$. So we have again $lim_{b to infty} 0 = 0$
Any comment about what I wrote would be really appreciated!
complex-analysis proof-verification contour-integration
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
add a comment |
Let $x > 0$. Find $$lim_{b to infty} int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt$$
I solved this problem but I am not quite confident about my steps. So please, could you check my solutions and tell me if there is something wrong??. Thanks in advanced!
First approuch:
Let's just focus on the first member of the integral:
$$int_{-b}^{b} frac{1}{t + ix} dt =
int_{-b}^{b} frac{1}{t + ix} Big( frac{t - ix}{t - ix} Big) dt =
int_{-b}^{b} frac{t - ix}{t^2 + x^2} dt$$
where the integrand can be again separated:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt - int_{-b}^{b} frac{ix}{t^2 + x^2}dt$$
Then, since the integral is about a real number $t$, $i$ can be treated as a constant. Now, I notice that the second integrand is an even function so the integral from $-b$ to $b$ is $0$. So we have:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt =
frac{1}{2}lnBig(t^2 + x^2 Big) Bigrvert_{-b}^{b} =
frac{1}{2}lnBig(frac{b^2 + x^2}{b^2 + x^2} Big) = 0$$
Following almost the same steps one can also show that $$int_{-b}^{b} frac{1}{t - ix} dt= 0$$
So the question reminds as $$lim_{b to infty} (0 + 0) = 0$$
Is this correct?? I mean:
Can $i$ be really treated as a constat in an integral which is about a real number??
Why would the question ask for about a limit int the infinite if the integral was destined to be $0$??
Second approuch
If $i$ can be really treated as a constant then the integral of the function $frac{1}{t + ix}$ would be $ln(t + ix)$, since $ix$ would be also a constant, so we have:
$$int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt=
ln(t + ix)Bigrvert_{-b}^{b} - ln(t - ix)Bigrvert_{-b}^{b}$$
Since $ix$ is just a constant then $ln$ fullfills the same properties as the well-known real valued function $ln$, so:
$$ = lnBig(frac{b + ix}{-b + ix} Big) - lnBig(frac{b - ix}{-b - ix} Big) =
lnBig(frac{frac{b + ix}{-b + ix}}{frac{b - ix}{-b - ix}} Big) = ln(1) = 0$$
Soo the answer would be $lim_{b to infty} 0 = 0$. I am really confused with theese steps since $ln$ function was evaluated in $b + ix$ and although I know that there is a counter-part complex version of the real function $ln$, I had accepted $ix$ as a costant, so there wouldn't be nothing wrong, right??
Third approuch
If $i$ can be really treated as a constant, then:
$$int_{-b}^{b} Big( frac{1}{t+ ix} - frac{1}{t - ix} Big) dt =
int_{-b}^{b} Big( frac{-2ix}{t^2 + x^2} Big) dt$$
because the integrand is an even function, that integral is $0$. So we have again $lim_{b to infty} 0 = 0$
Any comment about what I wrote would be really appreciated!
complex-analysis proof-verification contour-integration
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
Let $x > 0$. Find $$lim_{b to infty} int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt$$
I solved this problem but I am not quite confident about my steps. So please, could you check my solutions and tell me if there is something wrong??. Thanks in advanced!
First approuch:
Let's just focus on the first member of the integral:
$$int_{-b}^{b} frac{1}{t + ix} dt =
int_{-b}^{b} frac{1}{t + ix} Big( frac{t - ix}{t - ix} Big) dt =
int_{-b}^{b} frac{t - ix}{t^2 + x^2} dt$$
where the integrand can be again separated:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt - int_{-b}^{b} frac{ix}{t^2 + x^2}dt$$
Then, since the integral is about a real number $t$, $i$ can be treated as a constant. Now, I notice that the second integrand is an even function so the integral from $-b$ to $b$ is $0$. So we have:
$$ = int_{-b}^{b} frac{t}{t^2 + x^2}dt =
frac{1}{2}lnBig(t^2 + x^2 Big) Bigrvert_{-b}^{b} =
frac{1}{2}lnBig(frac{b^2 + x^2}{b^2 + x^2} Big) = 0$$
Following almost the same steps one can also show that $$int_{-b}^{b} frac{1}{t - ix} dt= 0$$
So the question reminds as $$lim_{b to infty} (0 + 0) = 0$$
Is this correct?? I mean:
Can $i$ be really treated as a constat in an integral which is about a real number??
Why would the question ask for about a limit int the infinite if the integral was destined to be $0$??
Second approuch
If $i$ can be really treated as a constant then the integral of the function $frac{1}{t + ix}$ would be $ln(t + ix)$, since $ix$ would be also a constant, so we have:
$$int_{-b}^{b} Big( frac{1}{t + ix} - frac{1}{t - ix} Big) dt=
ln(t + ix)Bigrvert_{-b}^{b} - ln(t - ix)Bigrvert_{-b}^{b}$$
Since $ix$ is just a constant then $ln$ fullfills the same properties as the well-known real valued function $ln$, so:
$$ = lnBig(frac{b + ix}{-b + ix} Big) - lnBig(frac{b - ix}{-b - ix} Big) =
lnBig(frac{frac{b + ix}{-b + ix}}{frac{b - ix}{-b - ix}} Big) = ln(1) = 0$$
Soo the answer would be $lim_{b to infty} 0 = 0$. I am really confused with theese steps since $ln$ function was evaluated in $b + ix$ and although I know that there is a counter-part complex version of the real function $ln$, I had accepted $ix$ as a costant, so there wouldn't be nothing wrong, right??
Third approuch
If $i$ can be really treated as a constant, then:
$$int_{-b}^{b} Big( frac{1}{t+ ix} - frac{1}{t - ix} Big) dt =
int_{-b}^{b} Big( frac{-2ix}{t^2 + x^2} Big) dt$$
because the integrand is an even function, that integral is $0$. So we have again $lim_{b to infty} 0 = 0$
Any comment about what I wrote would be really appreciated!
complex-analysis proof-verification contour-integration
complex-analysis proof-verification contour-integration
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
edited Nov 21 '18 at 11:21
Tianlalu
3,09621038
3,09621038
asked Nov 21 '18 at 11:15
Aldebaran
284
asked Nov 21 '18 at 11:15
Aldebaran
284
asked Nov 21 '18 at 11:15
Aldebaran
284
284
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $int_{-b}^b f(x) dx=2int_0^b f(x) dx$.
The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2kpi i, kinmathbb{Z}$.
answered Nov 21 '18 at 12:31
MoKo19
1914
add a comment |
If $x=0$, then the integral vanishes. Assume that $xne 0$. Then for $x>0$
$$
int_{-b}^bleft(frac{1}{t+ix}-frac{1}{t-ix}right),dt=
int_{-b}^bfrac{-2ix,dt}{t^2+x^2}=int_{-b}^bfrac{-2i,d(t/x)}{(t/x)^2+1}
=-2iint_{-b/x}^{b/x}frac{ds}{s^2+1}=-2i big(tan^{-1}(b/x)-tan^{-1}(-b/x)big)=-4itan^{-1}(b/x)to -4icdotfrac{pi}{2}=-2pi i,
$$
For $x<0$, the limit is $2pi i$.
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
add a comment |
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2 Answers
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active
oldest
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2 Answers
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active
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oldest
votes
The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $int_{-b}^b f(x) dx=2int_0^b f(x) dx$.
The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2kpi i, kinmathbb{Z}$.
answered Nov 21 '18 at 12:31
MoKo19
1914
add a comment |
The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $int_{-b}^b f(x) dx=2int_0^b f(x) dx$.
The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2kpi i, kinmathbb{Z}$.
answered Nov 21 '18 at 12:31
MoKo19
1914
add a comment |
The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $int_{-b}^b f(x) dx=2int_0^b f(x) dx$.
The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2kpi i, kinmathbb{Z}$.
answered Nov 21 '18 at 12:31
MoKo19
1914
The problem with your first and third approaches is that you wrote that any even function $f(x)$, we get that $int_{-b}^b f(x) dx=0$. This statement is true for odd functions. For even functions, $int_{-b}^b f(x) dx=2int_0^b f(x) dx$.
The problem with your second approach is that in the world of complex functions, logarithms are multiple-valued, and as such give you an answer that is true up to a $+2kpi i, kinmathbb{Z}$.
answered Nov 21 '18 at 12:31
MoKo19
1914
answered Nov 21 '18 at 12:31
MoKo19
1914
answered Nov 21 '18 at 12:31
MoKo19
1914
answered Nov 21 '18 at 12:31
MoKo19
1914
1914
add a comment |
add a comment |
If $x=0$, then the integral vanishes. Assume that $xne 0$. Then for $x>0$
$$
int_{-b}^bleft(frac{1}{t+ix}-frac{1}{t-ix}right),dt=
int_{-b}^bfrac{-2ix,dt}{t^2+x^2}=int_{-b}^bfrac{-2i,d(t/x)}{(t/x)^2+1}
=-2iint_{-b/x}^{b/x}frac{ds}{s^2+1}=-2i big(tan^{-1}(b/x)-tan^{-1}(-b/x)big)=-4itan^{-1}(b/x)to -4icdotfrac{pi}{2}=-2pi i,
$$
For $x<0$, the limit is $2pi i$.
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
add a comment |
If $x=0$, then the integral vanishes. Assume that $xne 0$. Then for $x>0$
$$
int_{-b}^bleft(frac{1}{t+ix}-frac{1}{t-ix}right),dt=
int_{-b}^bfrac{-2ix,dt}{t^2+x^2}=int_{-b}^bfrac{-2i,d(t/x)}{(t/x)^2+1}
=-2iint_{-b/x}^{b/x}frac{ds}{s^2+1}=-2i big(tan^{-1}(b/x)-tan^{-1}(-b/x)big)=-4itan^{-1}(b/x)to -4icdotfrac{pi}{2}=-2pi i,
$$
For $x<0$, the limit is $2pi i$.
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
add a comment |
If $x=0$, then the integral vanishes. Assume that $xne 0$. Then for $x>0$
$$
int_{-b}^bleft(frac{1}{t+ix}-frac{1}{t-ix}right),dt=
int_{-b}^bfrac{-2ix,dt}{t^2+x^2}=int_{-b}^bfrac{-2i,d(t/x)}{(t/x)^2+1}
=-2iint_{-b/x}^{b/x}frac{ds}{s^2+1}=-2i big(tan^{-1}(b/x)-tan^{-1}(-b/x)big)=-4itan^{-1}(b/x)to -4icdotfrac{pi}{2}=-2pi i,
$$
For $x<0$, the limit is $2pi i$.
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
If $x=0$, then the integral vanishes. Assume that $xne 0$. Then for $x>0$
$$
int_{-b}^bleft(frac{1}{t+ix}-frac{1}{t-ix}right),dt=
int_{-b}^bfrac{-2ix,dt}{t^2+x^2}=int_{-b}^bfrac{-2i,d(t/x)}{(t/x)^2+1}
=-2iint_{-b/x}^{b/x}frac{ds}{s^2+1}=-2i big(tan^{-1}(b/x)-tan^{-1}(-b/x)big)=-4itan^{-1}(b/x)to -4icdotfrac{pi}{2}=-2pi i,
$$
For $x<0$, the limit is $2pi i$.
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
answered Nov 21 '18 at 11:33
Yiorgos S. Smyrlis
62.8k1383163
62.8k1383163
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
add a comment |
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
wao. thank you so much, but where was I wrong to get a different answer¡?¡?
– Aldebaran
Nov 21 '18 at 12:03
add a comment |
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