Mixing
Solving a real integral in the complex plane
$begingroup$
$int_0^infty frac{cos(x)}{x^2+1}dx$
Singularities: $x_{1,2}=pm i$
We want to integrate over the upper half of a circle on the complex plane. So we only consider $x_1=+i$. We can use the residue theorem.
Define $h(z)=cos(z), g(z)=x^2+1, g'(z)=2x$
Since $g(i)=0, g'(i)=2ineq 0$ and $h()=cos(i)neq 0$ we can conclude that $x_1$ is a simple pole and we can use:
$Res(f;x_1)=frac{h(x_1}{g'(x_1)}=frac{cos(i)}{2i}=frac{e^{-1}+e^1}{4i}$
Now, since the integral is from $0$ to $infty$ we can conclude that:
$int_0^infty f(x)dx=pi i cdot (frac{e^{-1}+e^1}{4i})=frac{pi}{4}(e^{-1}+e^1)$
Now, I don't have any solutions but someone told me the solution is probably $frac{pi}{2e}$, if that's right - what did I do wrong?
integration complex-analysis
asked Jan 29 at 19:47
xotixxotix
291411
$endgroup$
add a comment |
$begingroup$
$int_0^infty frac{cos(x)}{x^2+1}dx$
Singularities: $x_{1,2}=pm i$
We want to integrate over the upper half of a circle on the complex plane. So we only consider $x_1=+i$. We can use the residue theorem.
Define $h(z)=cos(z), g(z)=x^2+1, g'(z)=2x$
Since $g(i)=0, g'(i)=2ineq 0$ and $h()=cos(i)neq 0$ we can conclude that $x_1$ is a simple pole and we can use:
$Res(f;x_1)=frac{h(x_1}{g'(x_1)}=frac{cos(i)}{2i}=frac{e^{-1}+e^1}{4i}$
Now, since the integral is from $0$ to $infty$ we can conclude that:
$int_0^infty f(x)dx=pi i cdot (frac{e^{-1}+e^1}{4i})=frac{pi}{4}(e^{-1}+e^1)$
Now, I don't have any solutions but someone told me the solution is probably $frac{pi}{2e}$, if that's right - what did I do wrong?
integration complex-analysis
asked Jan 29 at 19:47
xotixxotix
291411
$endgroup$
$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
$endgroup$
– Lord Shark the Unknown
Jan 29 at 20:02
$begingroup$
You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
$endgroup$
– xotix
Jan 29 at 20:35
add a comment |
$begingroup$
$int_0^infty frac{cos(x)}{x^2+1}dx$
Singularities: $x_{1,2}=pm i$
We want to integrate over the upper half of a circle on the complex plane. So we only consider $x_1=+i$. We can use the residue theorem.
Define $h(z)=cos(z), g(z)=x^2+1, g'(z)=2x$
Since $g(i)=0, g'(i)=2ineq 0$ and $h()=cos(i)neq 0$ we can conclude that $x_1$ is a simple pole and we can use:
$Res(f;x_1)=frac{h(x_1}{g'(x_1)}=frac{cos(i)}{2i}=frac{e^{-1}+e^1}{4i}$
Now, since the integral is from $0$ to $infty$ we can conclude that:
$int_0^infty f(x)dx=pi i cdot (frac{e^{-1}+e^1}{4i})=frac{pi}{4}(e^{-1}+e^1)$
Now, I don't have any solutions but someone told me the solution is probably $frac{pi}{2e}$, if that's right - what did I do wrong?
integration complex-analysis
asked Jan 29 at 19:47
xotixxotix
291411
$endgroup$
$int_0^infty frac{cos(x)}{x^2+1}dx$
Singularities: $x_{1,2}=pm i$
We want to integrate over the upper half of a circle on the complex plane. So we only consider $x_1=+i$. We can use the residue theorem.
Define $h(z)=cos(z), g(z)=x^2+1, g'(z)=2x$
Since $g(i)=0, g'(i)=2ineq 0$ and $h()=cos(i)neq 0$ we can conclude that $x_1$ is a simple pole and we can use:
$Res(f;x_1)=frac{h(x_1}{g'(x_1)}=frac{cos(i)}{2i}=frac{e^{-1}+e^1}{4i}$
Now, since the integral is from $0$ to $infty$ we can conclude that:
$int_0^infty f(x)dx=pi i cdot (frac{e^{-1}+e^1}{4i})=frac{pi}{4}(e^{-1}+e^1)$
Now, I don't have any solutions but someone told me the solution is probably $frac{pi}{2e}$, if that's right - what did I do wrong?
integration complex-analysis
integration complex-analysis
asked Jan 29 at 19:47
xotixxotix
291411
asked Jan 29 at 19:47
xotixxotix
291411
asked Jan 29 at 19:47
xotixxotix
291411
asked Jan 29 at 19:47
xotixxotix
291411
asked Jan 29 at 19:47
xotixxotix
291411
291411
$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
$endgroup$
– Lord Shark the Unknown
Jan 29 at 20:02
$begingroup$
You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
$endgroup$
– xotix
Jan 29 at 20:35
add a comment |
$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
$endgroup$
– Lord Shark the Unknown
Jan 29 at 20:02
$begingroup$
You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
$endgroup$
– xotix
Jan 29 at 20:35
$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
$endgroup$
– Lord Shark the Unknown
Jan 29 at 20:02
$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
$endgroup$
– Lord Shark the Unknown
Jan 29 at 20:02
$begingroup$
You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
$endgroup$
– xotix
Jan 29 at 20:35
$begingroup$
You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
$endgroup$
– xotix
Jan 29 at 20:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all a general note: never use trigonometric functions in such exercises. They are very hard to bound on half a circle. This is where the problems in your solution started. So instead of that define the complex function $f(z)=frac{e^{iz}}{z^2+1}$. Because the pole at the point $z=i$ is simple you can find the residue like this:
$Res(f;i)=lim_{zto i} (z-i)f(z)=lim_{zto i}frac{e^{iz}}{z+i}=frac{e^{-1}}{2i}$
So if we call our contour (for a specific radius) $Gamma$ then $int_{Gamma}f(z)dz=frac{pi}{e}$ by the residue theorem. On the other hand if you take the radius of the half circle to infinity then you can easily get that on the circle itself the limit is zero while on on the part of the contour which is on the real line the limit equals to $int_{-infty}^infty f(t)dt$. Using the identity $e^{it}=cost+isint$ we can take the real part of the integral and get that $int_{-infty}^inftyfrac{cosx}{x^2+1}dx=frac{pi}{e}$. This is an even function so the integral from $0$ to $infty$ is half of that.
answered Jan 29 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
$endgroup$
– xotix
Jan 29 at 20:27
$begingroup$
The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
$endgroup$
– Mark
Jan 29 at 20:36
$begingroup$
So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
$endgroup$
– Mark
Jan 29 at 20:40
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
|
show 1 more comment
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1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all a general note: never use trigonometric functions in such exercises. They are very hard to bound on half a circle. This is where the problems in your solution started. So instead of that define the complex function $f(z)=frac{e^{iz}}{z^2+1}$. Because the pole at the point $z=i$ is simple you can find the residue like this:
$Res(f;i)=lim_{zto i} (z-i)f(z)=lim_{zto i}frac{e^{iz}}{z+i}=frac{e^{-1}}{2i}$
So if we call our contour (for a specific radius) $Gamma$ then $int_{Gamma}f(z)dz=frac{pi}{e}$ by the residue theorem. On the other hand if you take the radius of the half circle to infinity then you can easily get that on the circle itself the limit is zero while on on the part of the contour which is on the real line the limit equals to $int_{-infty}^infty f(t)dt$. Using the identity $e^{it}=cost+isint$ we can take the real part of the integral and get that $int_{-infty}^inftyfrac{cosx}{x^2+1}dx=frac{pi}{e}$. This is an even function so the integral from $0$ to $infty$ is half of that.
answered Jan 29 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
$endgroup$
– xotix
Jan 29 at 20:27
$begingroup$
The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
$endgroup$
– Mark
Jan 29 at 20:36
$begingroup$
So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
$endgroup$
– Mark
Jan 29 at 20:40
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
|
show 1 more comment
$begingroup$
First of all a general note: never use trigonometric functions in such exercises. They are very hard to bound on half a circle. This is where the problems in your solution started. So instead of that define the complex function $f(z)=frac{e^{iz}}{z^2+1}$. Because the pole at the point $z=i$ is simple you can find the residue like this:
$Res(f;i)=lim_{zto i} (z-i)f(z)=lim_{zto i}frac{e^{iz}}{z+i}=frac{e^{-1}}{2i}$
So if we call our contour (for a specific radius) $Gamma$ then $int_{Gamma}f(z)dz=frac{pi}{e}$ by the residue theorem. On the other hand if you take the radius of the half circle to infinity then you can easily get that on the circle itself the limit is zero while on on the part of the contour which is on the real line the limit equals to $int_{-infty}^infty f(t)dt$. Using the identity $e^{it}=cost+isint$ we can take the real part of the integral and get that $int_{-infty}^inftyfrac{cosx}{x^2+1}dx=frac{pi}{e}$. This is an even function so the integral from $0$ to $infty$ is half of that.
answered Jan 29 at 20:01
MarkMark
10.4k1622
$endgroup$
$begingroup$
What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
$endgroup$
– xotix
Jan 29 at 20:27
$begingroup$
The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
$endgroup$
– Mark
Jan 29 at 20:36
$begingroup$
So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
$endgroup$
– Mark
Jan 29 at 20:40
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
|
show 1 more comment
$begingroup$
First of all a general note: never use trigonometric functions in such exercises. They are very hard to bound on half a circle. This is where the problems in your solution started. So instead of that define the complex function $f(z)=frac{e^{iz}}{z^2+1}$. Because the pole at the point $z=i$ is simple you can find the residue like this:
$Res(f;i)=lim_{zto i} (z-i)f(z)=lim_{zto i}frac{e^{iz}}{z+i}=frac{e^{-1}}{2i}$
So if we call our contour (for a specific radius) $Gamma$ then $int_{Gamma}f(z)dz=frac{pi}{e}$ by the residue theorem. On the other hand if you take the radius of the half circle to infinity then you can easily get that on the circle itself the limit is zero while on on the part of the contour which is on the real line the limit equals to $int_{-infty}^infty f(t)dt$. Using the identity $e^{it}=cost+isint$ we can take the real part of the integral and get that $int_{-infty}^inftyfrac{cosx}{x^2+1}dx=frac{pi}{e}$. This is an even function so the integral from $0$ to $infty$ is half of that.
answered Jan 29 at 20:01
MarkMark
10.4k1622
$endgroup$
First of all a general note: never use trigonometric functions in such exercises. They are very hard to bound on half a circle. This is where the problems in your solution started. So instead of that define the complex function $f(z)=frac{e^{iz}}{z^2+1}$. Because the pole at the point $z=i$ is simple you can find the residue like this:
$Res(f;i)=lim_{zto i} (z-i)f(z)=lim_{zto i}frac{e^{iz}}{z+i}=frac{e^{-1}}{2i}$
So if we call our contour (for a specific radius) $Gamma$ then $int_{Gamma}f(z)dz=frac{pi}{e}$ by the residue theorem. On the other hand if you take the radius of the half circle to infinity then you can easily get that on the circle itself the limit is zero while on on the part of the contour which is on the real line the limit equals to $int_{-infty}^infty f(t)dt$. Using the identity $e^{it}=cost+isint$ we can take the real part of the integral and get that $int_{-infty}^inftyfrac{cosx}{x^2+1}dx=frac{pi}{e}$. This is an even function so the integral from $0$ to $infty$ is half of that.
answered Jan 29 at 20:01
MarkMark
10.4k1622
edited Jan 29 at 20:07
answered Jan 29 at 20:01
MarkMark
10.4k1622
answered Jan 29 at 20:01
MarkMark
10.4k1622
answered Jan 29 at 20:01
MarkMark
10.4k1622
10.4k1622
$begingroup$
What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
$endgroup$
– xotix
Jan 29 at 20:27
$begingroup$
The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
$endgroup$
– Mark
Jan 29 at 20:36
$begingroup$
So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
$endgroup$
– Mark
Jan 29 at 20:40
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
|
show 1 more comment
$begingroup$
What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
$endgroup$
– xotix
Jan 29 at 20:27
$begingroup$
The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
$endgroup$
– Mark
Jan 29 at 20:36
$begingroup$
So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
$endgroup$
– Mark
Jan 29 at 20:40
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Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
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– xotix
Jan 29 at 20:56
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Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
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– Mark
Jan 29 at 21:02
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What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
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– xotix
Jan 29 at 20:27
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What I usually do is: I choose the upper half-circle with radius R in the complex plane as my path. Then I basically have $int_{-infty}^infty f(z)dz=int_{gamma_R}f(z)dz + int_{gamma_I} f(z) dz = 2pi sum_i Res(f;z_i)$. Then if the integral over the imaginary part of the path converges to 0, we can let $Rtoinfty$ and thus basically get the integral in $mathbb R$. Usually I only had functions, where the numerator is a simple function like $z^2$. So maybe the real question is: How does the function (in the numerator) affect this approach? You hinted to that. Mind explaining that more?
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– xotix
Jan 29 at 20:27
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The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
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– Mark
Jan 29 at 20:36
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The problem is that if you take your function to be $frac{cosz}{z^2+1}$ then the limit on the imaginary part will not be equal to $0$ when $Rtoinfty$. Trigonometric functions are very hard to bound. So we use the exponential function. If we define $f(z)=frac{e^{iz}}{z^2+1}$ like I did then the limit will be equal to zero. Write $z=x+iy$. Then $e^{iz}=e^{ix}e^{-y}$. The absolute value of $e^{ix}$ is $1$ and because we are in the upper half plane ($y>0$) we have $e^{-y}<1$. Hence $|e^{iz}|leq 1$ in the upper half plane. From here it is easy to conclude that the integral of $f$ goes to zero.
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– Mark
Jan 29 at 20:36
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So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
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– Mark
Jan 29 at 20:40
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So we find the integral $int_{-infty}^inftyfrac{e^{ix}}{x^2+1}dx$. And then using the identity $e^{ix}=cosx+isinx$ and the fact that we are on the real line we take the real part and get the required integral.
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– Mark
Jan 29 at 20:40
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Thanks a lot, that makes it way clearer! I also noticed that I didn't know about Jordan's Lemma, which you basically used here - np? Very nice, thanks!
$endgroup$
– xotix
Jan 29 at 20:56
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
$begingroup$
Actually I didn't use Jordan's lemma. I used the most simple bound: the absolute value of the integral is bounded by the length of the contour times the maximum of the function on this countour. So in our case it is bounded by $pi R*max|frac{e^{iz}}{z^2+1}|$. As we already know on this contour $|e^{iz}|leq 1$ and from here it is easy to see that $pi R*max|frac{e^{iz}}{z^2+1}|to 0$ when $Rtoinfty$. Now, if we would have an expression like $z-i$ in the denominator then this limit wouldn't be equal to $0$. In that case we would need Jordan's lemma. (so good to know it as well)
$endgroup$
– Mark
Jan 29 at 21:02
|
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$begingroup$
You didn't estimate the integral over the semicircular arc. (It doesn't tend to zero!)
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– Lord Shark the Unknown
Jan 29 at 20:02
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You are right, I didn't. Haven't done this in a long time, forgot about it! I usually never had to actually estimate it but I could argue differently. Im gonna check that, thanks.
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– xotix
Jan 29 at 20:35