Evaluate the integrals $int_gammaz^ndz$ for all integers $n$.
Clash Royale CLAN TAG#URR8PPP
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Evaluate the integrals
$$int_gammaz^ndz$$
for all integers $n$. Here $gamma$ is any circle centered at the origin with the positive (counterclockwise) orientation.
I don't know how to proceed. I couldn't find a "quick" way to do this, so I thought I'd use induction over $n$, but it seems like unnecessary work. Is there a clever way to calculate this integral?
My attempt. $z(t) = re^it$, so
$$int_gammaz^ndz = int_0^2pi(re^it)^nire^itdt = ir^n+1int_0^2pi(e^it)^n+1dt.$$
complex-analysis contour-integration
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
add a comment |Â
up vote
1
down vote
favorite
Evaluate the integrals
$$int_gammaz^ndz$$
for all integers $n$. Here $gamma$ is any circle centered at the origin with the positive (counterclockwise) orientation.
I don't know how to proceed. I couldn't find a "quick" way to do this, so I thought I'd use induction over $n$, but it seems like unnecessary work. Is there a clever way to calculate this integral?
My attempt. $z(t) = re^it$, so
$$int_gammaz^ndz = int_0^2pi(re^it)^nire^itdt = ir^n+1int_0^2pi(e^it)^n+1dt.$$
complex-analysis contour-integration
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Evaluate the integrals
$$int_gammaz^ndz$$
for all integers $n$. Here $gamma$ is any circle centered at the origin with the positive (counterclockwise) orientation.
I don't know how to proceed. I couldn't find a "quick" way to do this, so I thought I'd use induction over $n$, but it seems like unnecessary work. Is there a clever way to calculate this integral?
My attempt. $z(t) = re^it$, so
$$int_gammaz^ndz = int_0^2pi(re^it)^nire^itdt = ir^n+1int_0^2pi(e^it)^n+1dt.$$
complex-analysis contour-integration
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
Evaluate the integrals
$$int_gammaz^ndz$$
for all integers $n$. Here $gamma$ is any circle centered at the origin with the positive (counterclockwise) orientation.
I don't know how to proceed. I couldn't find a "quick" way to do this, so I thought I'd use induction over $n$, but it seems like unnecessary work. Is there a clever way to calculate this integral?
My attempt. $z(t) = re^it$, so
$$int_gammaz^ndz = int_0^2pi(re^it)^nire^itdt = ir^n+1int_0^2pi(e^it)^n+1dt.$$
complex-analysis contour-integration
complex-analysis contour-integration
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
asked Sep 7 at 3:01
Lucas Corrêa
1,133319
1,133319
Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26
add a comment |Â
Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26
Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26
add a comment |Â
1 Answer
1
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oldest
votes
up vote
1
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Your work looks perfectly good to me, you just need to finish it. If $n neq -1$, then $n+1 neq 0$, so $int_0^2 pi (e^it)^n+1 = 0$ (why?) If $n=-1$, then your reasoning shows the result is $ i int_0^2 pi 1 dz = 2 pi i $.
answered Sep 7 at 3:25
msm
970515
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Your work looks perfectly good to me, you just need to finish it. If $n neq -1$, then $n+1 neq 0$, so $int_0^2 pi (e^it)^n+1 = 0$ (why?) If $n=-1$, then your reasoning shows the result is $ i int_0^2 pi 1 dz = 2 pi i $.
answered Sep 7 at 3:25
msm
970515
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
add a comment |Â
up vote
1
down vote
accepted
Your work looks perfectly good to me, you just need to finish it. If $n neq -1$, then $n+1 neq 0$, so $int_0^2 pi (e^it)^n+1 = 0$ (why?) If $n=-1$, then your reasoning shows the result is $ i int_0^2 pi 1 dz = 2 pi i $.
answered Sep 7 at 3:25
msm
970515
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Your work looks perfectly good to me, you just need to finish it. If $n neq -1$, then $n+1 neq 0$, so $int_0^2 pi (e^it)^n+1 = 0$ (why?) If $n=-1$, then your reasoning shows the result is $ i int_0^2 pi 1 dz = 2 pi i $.
answered Sep 7 at 3:25
msm
970515
Your work looks perfectly good to me, you just need to finish it. If $n neq -1$, then $n+1 neq 0$, so $int_0^2 pi (e^it)^n+1 = 0$ (why?) If $n=-1$, then your reasoning shows the result is $ i int_0^2 pi 1 dz = 2 pi i $.
answered Sep 7 at 3:25
msm
970515
answered Sep 7 at 3:25
msm
970515
answered Sep 7 at 3:25
msm
970515
answered Sep 7 at 3:25
msm
970515
970515
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
add a comment |Â
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
I think I got it! If $n neq -1$ so, $int_0^2pi(e^it)^n+1dt = frace^it(n+1)i(n+1)|_0^2pi$, right?
– Lucas Corrêa
Sep 7 at 3:31
Yeah, that looks good to me
– msm
Sep 7 at 4:00
Yeah, that looks good to me
– msm
Sep 7 at 4:00
add a comment |Â
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Hint: for positive $n$, the integral is zero because Cauchy.
– Sean Roberson
Sep 7 at 3:08
Now find the antiderivative. The case $n = -1$ is special.
– Hans Engler
Sep 7 at 3:08
@SeanRoberson, it's true, but I still cannot use this
– Lucas Corrêa
Sep 7 at 3:23
@HansEngler, $fracddtleft(frace^it(n+1)i(n+1)right) = frace^it(n+1)i(n+1)i(n+1) = (e^it)^n+1$, right? If $n neq -1$
– Lucas Corrêa
Sep 7 at 3:26