How to find the Fourier transform of $frac{(t^2+2)}{(t^4+4)}$ and use it to evaluate...
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1
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I'm asked to evaluate the Fourier transform of
$dfrac{(t^2+2)}{(t^4+4)}$, and then use it to evaluate the integral from minus infinity to plus infinity of
$dfrac{(t^2+2)^2}{(t^4+4)^2}$.
Part 1:
So using the residue theorem and integrating along the upper half circle,
I get residues at $(1+i)$ and $(-1+i)$.
Evaluating the residues, I get $$frac{-ie^{omega(1-i)}}{4}$$ and $$frac{-ie^{omega(1+i)}}{4}$$
I use the ML inequality to prove that the arc length goes to zero as $R$ (the radius) tends toward infinity.
Using the Residue Theorem, I get $$frac{pi * e^{omega} *(e^{i omega}+ e^{i omega})}{2}$$ which could be turned into $$pi*e^{omega}*cos(omega)$$
Part 2: For the second part, Plancherel's theorem is obviously of use. The problem is that the integral from minus infinity to plus infinity of $$pi*e^{omega}*cos(omega)$$ doesn't seem to converge, so I don't see if I made a mistake or forgot something somewhere, and if that's not the case, how I should proceed further.
Thanks for your help !
Edit: I forgot to mention that it's the Fourier transform of $frac{(t^2+2)}{(t^4+4)}$ that I need to find, not "just" the integral from minus to plus infinity. Sorry !
Edit 2: corrected an error in the calculation of the fourier transform. There was an extra 2
integration complex-analysis contour-integration fourier-transform residue-calculus
add a comment |
up vote
1
down vote
favorite
I'm asked to evaluate the Fourier transform of
$dfrac{(t^2+2)}{(t^4+4)}$, and then use it to evaluate the integral from minus infinity to plus infinity of
$dfrac{(t^2+2)^2}{(t^4+4)^2}$.
Part 1:
So using the residue theorem and integrating along the upper half circle,
I get residues at $(1+i)$ and $(-1+i)$.
Evaluating the residues, I get $$frac{-ie^{omega(1-i)}}{4}$$ and $$frac{-ie^{omega(1+i)}}{4}$$
I use the ML inequality to prove that the arc length goes to zero as $R$ (the radius) tends toward infinity.
Using the Residue Theorem, I get $$frac{pi * e^{omega} *(e^{i omega}+ e^{i omega})}{2}$$ which could be turned into $$pi*e^{omega}*cos(omega)$$
Part 2: For the second part, Plancherel's theorem is obviously of use. The problem is that the integral from minus infinity to plus infinity of $$pi*e^{omega}*cos(omega)$$ doesn't seem to converge, so I don't see if I made a mistake or forgot something somewhere, and if that's not the case, how I should proceed further.
Thanks for your help !
Edit: I forgot to mention that it's the Fourier transform of $frac{(t^2+2)}{(t^4+4)}$ that I need to find, not "just" the integral from minus to plus infinity. Sorry !
Edit 2: corrected an error in the calculation of the fourier transform. There was an extra 2
integration complex-analysis contour-integration fourier-transform residue-calculus
I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm asked to evaluate the Fourier transform of
$dfrac{(t^2+2)}{(t^4+4)}$, and then use it to evaluate the integral from minus infinity to plus infinity of
$dfrac{(t^2+2)^2}{(t^4+4)^2}$.
Part 1:
So using the residue theorem and integrating along the upper half circle,
I get residues at $(1+i)$ and $(-1+i)$.
Evaluating the residues, I get $$frac{-ie^{omega(1-i)}}{4}$$ and $$frac{-ie^{omega(1+i)}}{4}$$
I use the ML inequality to prove that the arc length goes to zero as $R$ (the radius) tends toward infinity.
Using the Residue Theorem, I get $$frac{pi * e^{omega} *(e^{i omega}+ e^{i omega})}{2}$$ which could be turned into $$pi*e^{omega}*cos(omega)$$
Part 2: For the second part, Plancherel's theorem is obviously of use. The problem is that the integral from minus infinity to plus infinity of $$pi*e^{omega}*cos(omega)$$ doesn't seem to converge, so I don't see if I made a mistake or forgot something somewhere, and if that's not the case, how I should proceed further.
Thanks for your help !
Edit: I forgot to mention that it's the Fourier transform of $frac{(t^2+2)}{(t^4+4)}$ that I need to find, not "just" the integral from minus to plus infinity. Sorry !
Edit 2: corrected an error in the calculation of the fourier transform. There was an extra 2
integration complex-analysis contour-integration fourier-transform residue-calculus
I'm asked to evaluate the Fourier transform of
$dfrac{(t^2+2)}{(t^4+4)}$, and then use it to evaluate the integral from minus infinity to plus infinity of
$dfrac{(t^2+2)^2}{(t^4+4)^2}$.
Part 1:
So using the residue theorem and integrating along the upper half circle,
I get residues at $(1+i)$ and $(-1+i)$.
Evaluating the residues, I get $$frac{-ie^{omega(1-i)}}{4}$$ and $$frac{-ie^{omega(1+i)}}{4}$$
I use the ML inequality to prove that the arc length goes to zero as $R$ (the radius) tends toward infinity.
Using the Residue Theorem, I get $$frac{pi * e^{omega} *(e^{i omega}+ e^{i omega})}{2}$$ which could be turned into $$pi*e^{omega}*cos(omega)$$
Part 2: For the second part, Plancherel's theorem is obviously of use. The problem is that the integral from minus infinity to plus infinity of $$pi*e^{omega}*cos(omega)$$ doesn't seem to converge, so I don't see if I made a mistake or forgot something somewhere, and if that's not the case, how I should proceed further.
Thanks for your help !
Edit: I forgot to mention that it's the Fourier transform of $frac{(t^2+2)}{(t^4+4)}$ that I need to find, not "just" the integral from minus to plus infinity. Sorry !
Edit 2: corrected an error in the calculation of the fourier transform. There was an extra 2
integration complex-analysis contour-integration fourier-transform residue-calculus
integration complex-analysis contour-integration fourier-transform residue-calculus
edited 11 hours ago
asked yesterday
Poujh
307212
307212
I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday
add a comment |
I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday
I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday
I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday
add a comment |
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
First: note that the approach you have taken is partial. Since this expression scales like $e^{-iomega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=pm1-i.$ This analysis switches for negative $omega$ so that you shrink to loops around $t=pm1+i.$
Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $omega$ and thus describes only an exponential decay; the expression for positive $omega$ has an $e^{-omega}.$
Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$int_{mathbb R} dt~e^{-iomega t}~frac1{1+t^2}=pi~e^{-|omega|},$$ and practice from there.
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First: note that the approach you have taken is partial. Since this expression scales like $e^{-iomega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=pm1-i.$ This analysis switches for negative $omega$ so that you shrink to loops around $t=pm1+i.$
Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $omega$ and thus describes only an exponential decay; the expression for positive $omega$ has an $e^{-omega}.$
Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$int_{mathbb R} dt~e^{-iomega t}~frac1{1+t^2}=pi~e^{-|omega|},$$ and practice from there.
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
add a comment |
up vote
2
down vote
accepted
First: note that the approach you have taken is partial. Since this expression scales like $e^{-iomega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=pm1-i.$ This analysis switches for negative $omega$ so that you shrink to loops around $t=pm1+i.$
Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $omega$ and thus describes only an exponential decay; the expression for positive $omega$ has an $e^{-omega}.$
Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$int_{mathbb R} dt~e^{-iomega t}~frac1{1+t^2}=pi~e^{-|omega|},$$ and practice from there.
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First: note that the approach you have taken is partial. Since this expression scales like $e^{-iomega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=pm1-i.$ This analysis switches for negative $omega$ so that you shrink to loops around $t=pm1+i.$
Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $omega$ and thus describes only an exponential decay; the expression for positive $omega$ has an $e^{-omega}.$
Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$int_{mathbb R} dt~e^{-iomega t}~frac1{1+t^2}=pi~e^{-|omega|},$$ and practice from there.
First: note that the approach you have taken is partial. Since this expression scales like $e^{-iomega t}/t^2,$ whether you can add a semicircle contour at infinity that integrates to zero depends on the sign of $omega.$ If it is positive then we need to add a contour on the negative imaginary half-plane to get $(-i)^2<0$ to avoid exponential blowup; shrinking the curve we have loops about $t=pm1-i.$ This analysis switches for negative $omega$ so that you shrink to loops around $t=pm1+i.$
Second: note that this will solve your convergence concerns as the expression you have is only valid for negative $omega$ and thus describes only an exponential decay; the expression for positive $omega$ has an $e^{-omega}.$
Third: if the above is still unclear try to start from an easier place: show that the Lorentzian $$int_{mathbb R} dt~e^{-iomega t}~frac1{1+t^2}=pi~e^{-|omega|},$$ and practice from there.
answered yesterday
CR Drost
1,800711
1,800711
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
add a comment |
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Thanks a lot ! I just have one question left. When I evaluate $dfrac{(t^2+2)^2}{(t^4+4)^2}$ in Wolfram Alpha, I get 3pi/8. But when I use Plancherel theorem, I only get the same result when I use $$pi*e^{omega}*cos(omega)$$, so without the factor of 2. It seems I added an extra 2 somewhere, but I checked my calculations several times and they seem to be correct. Do you know where this extra factor of 2 could come from ?
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Could it be that we multiply only with $pi*i$ and not $2*pi*i$ when using the residue theorem? But I don't see why this would be the case.
– Poujh
13 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
Nevermind. Found the error. It has nothing to do with the 2*pi*i. When we add both residues, we get $frac{-i*e^{omega(1+i)}}{4} + frac{-i*e^{omega(1-i)}}{4}$. My error was to not see that we cannot add both exponentials and thus cancel with a two in the denominator (to get $frac{-i(e^{omega(1+i)}+e^{omega(1-i)})}{2}$ which is wrong, there should be a 4 in the denominator because the two exponentials aren't the same. ) We can only factor out the $-i$ wich doesn't change anything in the denominator.
– Poujh
11 hours ago
add a comment |
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I meant the fourier transform of $frac{(t^2+2)}{(t^4+4)}$ , sorry !
– Poujh
yesterday