Mixing
Integration over a circle using (Cauchy's) integral formula
Clash Royale CLAN TAG#URR8PPP
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$$int_C[0,3]^fracsin(z)(z^2+1/2)^2,dz$$ is the integral I need to calculate.
I separated the integrand:
$$frac1bigl(z-frac1sqrt2ibigr)^2cdotfracsin(z)bigl(z+frac1sqrt2ibigr)^2$$
Consequently $$f(z)=fracsin(z)bigl(z+frac1sqrt2ibigr)^2$$ $$f'(z)=fraccos(z) bigl(z+frac1sqrt2ibigr)^2-2bigl(z+frac1sqrt2ibigr)sin(z)bigl(z+frac1sqrt2ibigr)^4$$ and I use the integral formula $$int_C[0,3]^fracsin(z)bigl(z^2+1/2bigr)^2,dz = 2pi i f'(w),enspace w=frac12i$$.
I got $$f'Bigl(frac1sqrt2 iBigr)=-frac12 coshBigl(frac1sqrt2Bigr)-frac1sqrt2 sinhBigl(frac1sqrt2Bigr)$$ but this is wrong. I don't know why; I hope one of you can help me.
the answer is $int_C[0,3]fracsin(z)(z^2+1/2)^2,dz = 2 pi isqrt2 sinhbigl(frac1sqrt2bigr)$.
complex-analysis
asked Jul 18 at 15:13
123
515
add a comment |Â
up vote
0
down vote
favorite
$$int_C[0,3]^fracsin(z)(z^2+1/2)^2,dz$$ is the integral I need to calculate.
I separated the integrand:
$$frac1bigl(z-frac1sqrt2ibigr)^2cdotfracsin(z)bigl(z+frac1sqrt2ibigr)^2$$
Consequently $$f(z)=fracsin(z)bigl(z+frac1sqrt2ibigr)^2$$ $$f'(z)=fraccos(z) bigl(z+frac1sqrt2ibigr)^2-2bigl(z+frac1sqrt2ibigr)sin(z)bigl(z+frac1sqrt2ibigr)^4$$ and I use the integral formula $$int_C[0,3]^fracsin(z)bigl(z^2+1/2bigr)^2,dz = 2pi i f'(w),enspace w=frac12i$$.
I got $$f'Bigl(frac1sqrt2 iBigr)=-frac12 coshBigl(frac1sqrt2Bigr)-frac1sqrt2 sinhBigl(frac1sqrt2Bigr)$$ but this is wrong. I don't know why; I hope one of you can help me.
the answer is $int_C[0,3]fracsin(z)(z^2+1/2)^2,dz = 2 pi isqrt2 sinhbigl(frac1sqrt2bigr)$.
complex-analysis
asked Jul 18 at 15:13
123
515
$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$$int_C[0,3]^fracsin(z)(z^2+1/2)^2,dz$$ is the integral I need to calculate.
I separated the integrand:
$$frac1bigl(z-frac1sqrt2ibigr)^2cdotfracsin(z)bigl(z+frac1sqrt2ibigr)^2$$
Consequently $$f(z)=fracsin(z)bigl(z+frac1sqrt2ibigr)^2$$ $$f'(z)=fraccos(z) bigl(z+frac1sqrt2ibigr)^2-2bigl(z+frac1sqrt2ibigr)sin(z)bigl(z+frac1sqrt2ibigr)^4$$ and I use the integral formula $$int_C[0,3]^fracsin(z)bigl(z^2+1/2bigr)^2,dz = 2pi i f'(w),enspace w=frac12i$$.
I got $$f'Bigl(frac1sqrt2 iBigr)=-frac12 coshBigl(frac1sqrt2Bigr)-frac1sqrt2 sinhBigl(frac1sqrt2Bigr)$$ but this is wrong. I don't know why; I hope one of you can help me.
the answer is $int_C[0,3]fracsin(z)(z^2+1/2)^2,dz = 2 pi isqrt2 sinhbigl(frac1sqrt2bigr)$.
complex-analysis
asked Jul 18 at 15:13
123
515
$$int_C[0,3]^fracsin(z)(z^2+1/2)^2,dz$$ is the integral I need to calculate.
I separated the integrand:
$$frac1bigl(z-frac1sqrt2ibigr)^2cdotfracsin(z)bigl(z+frac1sqrt2ibigr)^2$$
Consequently $$f(z)=fracsin(z)bigl(z+frac1sqrt2ibigr)^2$$ $$f'(z)=fraccos(z) bigl(z+frac1sqrt2ibigr)^2-2bigl(z+frac1sqrt2ibigr)sin(z)bigl(z+frac1sqrt2ibigr)^4$$ and I use the integral formula $$int_C[0,3]^fracsin(z)bigl(z^2+1/2bigr)^2,dz = 2pi i f'(w),enspace w=frac12i$$.
I got $$f'Bigl(frac1sqrt2 iBigr)=-frac12 coshBigl(frac1sqrt2Bigr)-frac1sqrt2 sinhBigl(frac1sqrt2Bigr)$$ but this is wrong. I don't know why; I hope one of you can help me.
the answer is $int_C[0,3]fracsin(z)(z^2+1/2)^2,dz = 2 pi isqrt2 sinhbigl(frac1sqrt2bigr)$.
complex-analysis
asked Jul 18 at 15:13
123
515
edited Jul 18 at 19:30
asked Jul 18 at 15:13
123
515
asked Jul 18 at 15:13
123
515
asked Jul 18 at 15:13
123
515
515
$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02
add a comment |Â
$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02
$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02
add a comment |Â
1 Answer
1
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oldest
votes
up vote
0
down vote
Hint: Try again with following decomposition
$$(z^2+dfrac12)=(z-dfracsqrt22i)(z+dfracsqrt22i)$$
answered Jul 18 at 15:21
Nosrati
19.6k41544
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Hint: Try again with following decomposition
$$(z^2+dfrac12)=(z-dfracsqrt22i)(z+dfracsqrt22i)$$
answered Jul 18 at 15:21
Nosrati
19.6k41544
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
add a comment |Â
up vote
0
down vote
Hint: Try again with following decomposition
$$(z^2+dfrac12)=(z-dfracsqrt22i)(z+dfracsqrt22i)$$
answered Jul 18 at 15:21
Nosrati
19.6k41544
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Hint: Try again with following decomposition
$$(z^2+dfrac12)=(z-dfracsqrt22i)(z+dfracsqrt22i)$$
answered Jul 18 at 15:21
Nosrati
19.6k41544
Hint: Try again with following decomposition
$$(z^2+dfrac12)=(z-dfracsqrt22i)(z+dfracsqrt22i)$$
answered Jul 18 at 15:21
Nosrati
19.6k41544
answered Jul 18 at 15:21
Nosrati
19.6k41544
answered Jul 18 at 15:21
Nosrati
19.6k41544
answered Jul 18 at 15:21
Nosrati
19.6k41544
19.6k41544
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
add a comment |Â
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
isn't that what I've done?
– 123
Jul 18 at 15:35
isn't that what I've done?
– 123
Jul 18 at 15:35
See radicals.!.
– Nosrati
Jul 18 at 15:42
See radicals.!.
– Nosrati
Jul 18 at 15:42
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
Aha, I had done that in my calculations but I edited my misstake now.
– 123
Jul 18 at 15:46
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
with what you wrote that produces the wrong answer, I think the problem lies elsewhere.
– 123
Jul 18 at 15:47
add a comment |Â
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$$f'(frac1sqrt2 i)=-frac12 cosh(frac1sqrt2)colorred+frac1sqrt2 sinh(frac1sqrt2)$$
– Nosrati
Jul 18 at 15:53
the answer is $f'(frac1sqrt2i) = sqrt2 sinh(frac1sqrt2)$ so the fault still lies somewhere else
– 123
Jul 18 at 16:02