Mixing
evaluate $int_0^{infty}frac{x^a}{x^2+1}dx$ for $ain(0,1)$
$begingroup$
I am stuck on this complex analysis homework problem :( here is my attack so far:
If I integrate around a keyhole contour, I can show that $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi i}{1-e^{2api i}}sum_{win{i,-i}}RES(frac{e^{alog z}}{z^2+1},w).$$ Now, luckily the poles in this integrand are simple, so it's easy enough to find: $$RES(frac{e^{alog z}}{z^2+1},i)=frac{e^{api i/2}}{2i},$$ while $$RES(frac{e^{alog z}}{z^2+1},-i)=frac{-e^{-api i/2}}{2i},$$ and adding these fellas together we get $sin(api/2)$ by Euler's i.d. So it goes $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi isin(api/2)}{1-e^{2api i}}.$$ But where to go now? The integral on the left is real, while I have my doubts about the term on the right.
Have I made a mistake, or there's some further trick I don't see? (I'm almost certain it's the former, since I'm an expert at mistakes but an amateur at finding them). Anyways, I would appreciate any help, or advice on this problem. Thanks!
contour-integration
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
$endgroup$
add a comment |
$begingroup$
I am stuck on this complex analysis homework problem :( here is my attack so far:
If I integrate around a keyhole contour, I can show that $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi i}{1-e^{2api i}}sum_{win{i,-i}}RES(frac{e^{alog z}}{z^2+1},w).$$ Now, luckily the poles in this integrand are simple, so it's easy enough to find: $$RES(frac{e^{alog z}}{z^2+1},i)=frac{e^{api i/2}}{2i},$$ while $$RES(frac{e^{alog z}}{z^2+1},-i)=frac{-e^{-api i/2}}{2i},$$ and adding these fellas together we get $sin(api/2)$ by Euler's i.d. So it goes $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi isin(api/2)}{1-e^{2api i}}.$$ But where to go now? The integral on the left is real, while I have my doubts about the term on the right.
Have I made a mistake, or there's some further trick I don't see? (I'm almost certain it's the former, since I'm an expert at mistakes but an amateur at finding them). Anyways, I would appreciate any help, or advice on this problem. Thanks!
contour-integration
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
$endgroup$
2
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39
add a comment |
$begingroup$
I am stuck on this complex analysis homework problem :( here is my attack so far:
If I integrate around a keyhole contour, I can show that $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi i}{1-e^{2api i}}sum_{win{i,-i}}RES(frac{e^{alog z}}{z^2+1},w).$$ Now, luckily the poles in this integrand are simple, so it's easy enough to find: $$RES(frac{e^{alog z}}{z^2+1},i)=frac{e^{api i/2}}{2i},$$ while $$RES(frac{e^{alog z}}{z^2+1},-i)=frac{-e^{-api i/2}}{2i},$$ and adding these fellas together we get $sin(api/2)$ by Euler's i.d. So it goes $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi isin(api/2)}{1-e^{2api i}}.$$ But where to go now? The integral on the left is real, while I have my doubts about the term on the right.
Have I made a mistake, or there's some further trick I don't see? (I'm almost certain it's the former, since I'm an expert at mistakes but an amateur at finding them). Anyways, I would appreciate any help, or advice on this problem. Thanks!
contour-integration
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
$endgroup$
I am stuck on this complex analysis homework problem :( here is my attack so far:
If I integrate around a keyhole contour, I can show that $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi i}{1-e^{2api i}}sum_{win{i,-i}}RES(frac{e^{alog z}}{z^2+1},w).$$ Now, luckily the poles in this integrand are simple, so it's easy enough to find: $$RES(frac{e^{alog z}}{z^2+1},i)=frac{e^{api i/2}}{2i},$$ while $$RES(frac{e^{alog z}}{z^2+1},-i)=frac{-e^{-api i/2}}{2i},$$ and adding these fellas together we get $sin(api/2)$ by Euler's i.d. So it goes $$int_0^{infty}frac{x^a}{x^2+1}dx=frac{2pi isin(api/2)}{1-e^{2api i}}.$$ But where to go now? The integral on the left is real, while I have my doubts about the term on the right.
Have I made a mistake, or there's some further trick I don't see? (I'm almost certain it's the former, since I'm an expert at mistakes but an amateur at finding them). Anyways, I would appreciate any help, or advice on this problem. Thanks!
contour-integration
contour-integration
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
asked Jan 26 at 22:03
T. AlexanderT. Alexander
273
273
2
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39
add a comment |
2
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39
2
2
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The trick to this one? You've got the wrong residues. We have to use the same branch of the function in calculating the residues that we did in setting up the integral. Specifically, since we run $arg z$ from $0$ to $2pi$, we should write the two zeros of the denominator as $i=expleft(frac{pi i}{2}right)$ and $-i=expleft(frac{3pi i}{2}right)$. The residues are then $frac1{2i}e^{api i/2}{2i}$ and $frac{-1}{2i}e^{3api i/2}$.
That leads to a factor of $e^{api i}$ in the numerator. Divide through by it, the denominator becomes $i$ times a sine, and it all comes out real.
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
$endgroup$
add a comment |
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$begingroup$
The trick to this one? You've got the wrong residues. We have to use the same branch of the function in calculating the residues that we did in setting up the integral. Specifically, since we run $arg z$ from $0$ to $2pi$, we should write the two zeros of the denominator as $i=expleft(frac{pi i}{2}right)$ and $-i=expleft(frac{3pi i}{2}right)$. The residues are then $frac1{2i}e^{api i/2}{2i}$ and $frac{-1}{2i}e^{3api i/2}$.
That leads to a factor of $e^{api i}$ in the numerator. Divide through by it, the denominator becomes $i$ times a sine, and it all comes out real.
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
$endgroup$
add a comment |
$begingroup$
The trick to this one? You've got the wrong residues. We have to use the same branch of the function in calculating the residues that we did in setting up the integral. Specifically, since we run $arg z$ from $0$ to $2pi$, we should write the two zeros of the denominator as $i=expleft(frac{pi i}{2}right)$ and $-i=expleft(frac{3pi i}{2}right)$. The residues are then $frac1{2i}e^{api i/2}{2i}$ and $frac{-1}{2i}e^{3api i/2}$.
That leads to a factor of $e^{api i}$ in the numerator. Divide through by it, the denominator becomes $i$ times a sine, and it all comes out real.
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
$endgroup$
add a comment |
$begingroup$
The trick to this one? You've got the wrong residues. We have to use the same branch of the function in calculating the residues that we did in setting up the integral. Specifically, since we run $arg z$ from $0$ to $2pi$, we should write the two zeros of the denominator as $i=expleft(frac{pi i}{2}right)$ and $-i=expleft(frac{3pi i}{2}right)$. The residues are then $frac1{2i}e^{api i/2}{2i}$ and $frac{-1}{2i}e^{3api i/2}$.
That leads to a factor of $e^{api i}$ in the numerator. Divide through by it, the denominator becomes $i$ times a sine, and it all comes out real.
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
$endgroup$
The trick to this one? You've got the wrong residues. We have to use the same branch of the function in calculating the residues that we did in setting up the integral. Specifically, since we run $arg z$ from $0$ to $2pi$, we should write the two zeros of the denominator as $i=expleft(frac{pi i}{2}right)$ and $-i=expleft(frac{3pi i}{2}right)$. The residues are then $frac1{2i}e^{api i/2}{2i}$ and $frac{-1}{2i}e^{3api i/2}$.
That leads to a factor of $e^{api i}$ in the numerator. Divide through by it, the denominator becomes $i$ times a sine, and it all comes out real.
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
answered Jan 26 at 22:24
jmerryjmerry
15.8k1632
15.8k1632
add a comment |
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2
$begingroup$
Factor out an $e^{pi a i}$ from the denominator
$endgroup$
– Frank W.
Jan 26 at 22:06
$begingroup$
@ Frank W. Oh I see! haha wow I'm dumb, thank you so much!!
$endgroup$
– T. Alexander
Jan 26 at 22:20
$begingroup$
While a useful technique, that doesn't actually solve the issue here. The expression written down is indeed not real.
$endgroup$
– jmerry
Jan 26 at 22:24
$begingroup$
@jmerry yes but it made it more clear what the issue with my solution was, also thank you for your answer as well!
$endgroup$
– T. Alexander
Jan 26 at 22:39