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Proof $Reint_-fracpi2^fracpi2frac1tan^2n(x)+1dx=fracpi2,forall n in mathbbR$ [duplicate]
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Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
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I was solving the integral $$int_-fracpi2^fracpi2frac1tan^6(x)+1dx$$ and here's my solution:
Let $I = int_-fracpi2^fracpi2frac1tan^6(x)+1dx$, then $$I=2int_0^fracpi2frac1tan^6(x)+1dx.$$ Subs. $u=tan x,dx=frac11+u^2du$.
$$I = 2int_0^inftyfrac1(1+u^6)(1+u^2)du$$
$$I=2left(int_0^1frac1(1+u^6)(1+u^2)du + int_1^inftyfrac1(1+u^6)(1+u^2)duright).$$
On the first integral, $u mapsto u^-1$, ending with $$I=2int_1^inftyfracdu1+u^2=fracpi2.$$
Using the same approach, I was able to prove that the equality holds for all $nin mathbbZ$. Going beyond, and using Wolfram Alpha, the equality seems to hold for all reals $n$, but only for the real part. So, here's my question:
Claim. Let $$mathcalI=Reint_-fracpi2^fracpi2frac1tan^2n(x)+1dx.$$ Then, $mathcalI=fracpi2, forall n in mathbbR.$
First of all: Is the Claim true?
If positive, can I prove it using the same methods?
calculus complex-analysis definite-integrals improper-integrals
asked Jul 30 at 22:42
otreblig
31129
marked as duplicate by Mark Viola calculus
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Jul 31 at 0:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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This question already has an answer here:
Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
3 answers
I was solving the integral $$int_-fracpi2^fracpi2frac1tan^6(x)+1dx$$ and here's my solution:
Let $I = int_-fracpi2^fracpi2frac1tan^6(x)+1dx$, then $$I=2int_0^fracpi2frac1tan^6(x)+1dx.$$ Subs. $u=tan x,dx=frac11+u^2du$.
$$I = 2int_0^inftyfrac1(1+u^6)(1+u^2)du$$
$$I=2left(int_0^1frac1(1+u^6)(1+u^2)du + int_1^inftyfrac1(1+u^6)(1+u^2)duright).$$
On the first integral, $u mapsto u^-1$, ending with $$I=2int_1^inftyfracdu1+u^2=fracpi2.$$
Using the same approach, I was able to prove that the equality holds for all $nin mathbbZ$. Going beyond, and using Wolfram Alpha, the equality seems to hold for all reals $n$, but only for the real part. So, here's my question:
Claim. Let $$mathcalI=Reint_-fracpi2^fracpi2frac1tan^2n(x)+1dx.$$ Then, $mathcalI=fracpi2, forall n in mathbbR.$
First of all: Is the Claim true?
If positive, can I prove it using the same methods?
calculus complex-analysis definite-integrals improper-integrals
asked Jul 30 at 22:42
otreblig
31129
marked as duplicate by Mark Viola calculus
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Jul 31 at 0:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53
add a comment |Â
up vote
1
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up vote
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down vote
favorite
This question already has an answer here:
Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
3 answers
I was solving the integral $$int_-fracpi2^fracpi2frac1tan^6(x)+1dx$$ and here's my solution:
Let $I = int_-fracpi2^fracpi2frac1tan^6(x)+1dx$, then $$I=2int_0^fracpi2frac1tan^6(x)+1dx.$$ Subs. $u=tan x,dx=frac11+u^2du$.
$$I = 2int_0^inftyfrac1(1+u^6)(1+u^2)du$$
$$I=2left(int_0^1frac1(1+u^6)(1+u^2)du + int_1^inftyfrac1(1+u^6)(1+u^2)duright).$$
On the first integral, $u mapsto u^-1$, ending with $$I=2int_1^inftyfracdu1+u^2=fracpi2.$$
Using the same approach, I was able to prove that the equality holds for all $nin mathbbZ$. Going beyond, and using Wolfram Alpha, the equality seems to hold for all reals $n$, but only for the real part. So, here's my question:
Claim. Let $$mathcalI=Reint_-fracpi2^fracpi2frac1tan^2n(x)+1dx.$$ Then, $mathcalI=fracpi2, forall n in mathbbR.$
First of all: Is the Claim true?
If positive, can I prove it using the same methods?
calculus complex-analysis definite-integrals improper-integrals
asked Jul 30 at 22:42
otreblig
31129
This question already has an answer here:
Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
3 answers
I was solving the integral $$int_-fracpi2^fracpi2frac1tan^6(x)+1dx$$ and here's my solution:
Let $I = int_-fracpi2^fracpi2frac1tan^6(x)+1dx$, then $$I=2int_0^fracpi2frac1tan^6(x)+1dx.$$ Subs. $u=tan x,dx=frac11+u^2du$.
$$I = 2int_0^inftyfrac1(1+u^6)(1+u^2)du$$
$$I=2left(int_0^1frac1(1+u^6)(1+u^2)du + int_1^inftyfrac1(1+u^6)(1+u^2)duright).$$
On the first integral, $u mapsto u^-1$, ending with $$I=2int_1^inftyfracdu1+u^2=fracpi2.$$
Using the same approach, I was able to prove that the equality holds for all $nin mathbbZ$. Going beyond, and using Wolfram Alpha, the equality seems to hold for all reals $n$, but only for the real part. So, here's my question:
Claim. Let $$mathcalI=Reint_-fracpi2^fracpi2frac1tan^2n(x)+1dx.$$ Then, $mathcalI=fracpi2, forall n in mathbbR.$
First of all: Is the Claim true?
If positive, can I prove it using the same methods?
This question already has an answer here:
Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
3 answers
calculus complex-analysis definite-integrals improper-integrals
asked Jul 30 at 22:42
otreblig
31129
asked Jul 30 at 22:42
otreblig
31129
asked Jul 30 at 22:42
otreblig
31129
asked Jul 30 at 22:42
otreblig
31129
31129
marked as duplicate by Mark Viola calculus
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Jul 31 at 0:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Mark Viola calculus
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Jul 31 at 0:02
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53
add a comment |Â
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53
add a comment |Â
2 Answers
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There will be no problem if you write $(tan^2 x)^n$. But you can try to insist on using the power $s:=2n$. If you were talking about $$2∫_0^pi/2fracdxtan^s x + 1$$ there would be no issue using your proof still, but the integral $$J:=∫_-pi/2^0fracdxtan^s x + 1 $$
can have an imaginary part. There could be one answer for each branch of $x^s=e^slog x $. Since $x<0$, these branches are
$$ e^x = |x|^s e^(2k+1)À i s,quad k ∈ mathbb Z$$
giving
$$ frac1tan^s x + 1 = frac1e^(2k+1)Ài s tan^s(-x) + 1$$
If you try to go through the same proof, you end up with
$$ J = ∫_0^1 fracv^s(v^s + e^(2k+1)iÀ s)(v^2+1) + frac1(e^(2k+1)iÀ sv^s + 1)(v^2+1) dv$$
One can check that
$$ fracv^2s exp(ix) + 2v^s + exp(ix)[v^s+exp(ix)][v^s exp(ix)+1] = 1 -2 i sin(x) (v^2s + 2 v^s cos(x) + 1)$$
with $x = (2k+1)À i s$, this verifies that regardless of the branch cut used to define $tan^s x$ for $x<0$,
$$Re∫_-pi/2^0 frac1tan^s x + 1 dx = fracÀ4 $$
answered Jul 31 at 0:23
Calvin Khor
8,00411132
add a comment |Â
up vote
0
down vote
Your integral always equals
$$2int^infty_0frac1(1+u^2n)(1+u^2)du$$
$$=2int^1_0 frac1(1+u^2n)(1+u^2)du +2int^infty_1 frac1(1+u^2n)(1+u^2)du$$
By substitution $umapsto frac1u$ on the second integral, one obtains
$$2int^1_0fracfrac1u^2du(1+1/u^2n)(1+1/u^2)=2int^1_0fracu^2n(u^2n+1)(1+u^2)du$$
Adding two integrals together, you see the cancellation and obtain
$$2int^1_0frac11+u^2du=fracpi2$$
So, yes, your method does work for any $n$.
answered Jul 30 at 23:22
Szeto
3,8431421
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
There will be no problem if you write $(tan^2 x)^n$. But you can try to insist on using the power $s:=2n$. If you were talking about $$2∫_0^pi/2fracdxtan^s x + 1$$ there would be no issue using your proof still, but the integral $$J:=∫_-pi/2^0fracdxtan^s x + 1 $$
can have an imaginary part. There could be one answer for each branch of $x^s=e^slog x $. Since $x<0$, these branches are
$$ e^x = |x|^s e^(2k+1)À i s,quad k ∈ mathbb Z$$
giving
$$ frac1tan^s x + 1 = frac1e^(2k+1)Ài s tan^s(-x) + 1$$
If you try to go through the same proof, you end up with
$$ J = ∫_0^1 fracv^s(v^s + e^(2k+1)iÀ s)(v^2+1) + frac1(e^(2k+1)iÀ sv^s + 1)(v^2+1) dv$$
One can check that
$$ fracv^2s exp(ix) + 2v^s + exp(ix)[v^s+exp(ix)][v^s exp(ix)+1] = 1 -2 i sin(x) (v^2s + 2 v^s cos(x) + 1)$$
with $x = (2k+1)À i s$, this verifies that regardless of the branch cut used to define $tan^s x$ for $x<0$,
$$Re∫_-pi/2^0 frac1tan^s x + 1 dx = fracÀ4 $$
answered Jul 31 at 0:23
Calvin Khor
8,00411132
add a comment |Â
up vote
1
down vote
accepted
There will be no problem if you write $(tan^2 x)^n$. But you can try to insist on using the power $s:=2n$. If you were talking about $$2∫_0^pi/2fracdxtan^s x + 1$$ there would be no issue using your proof still, but the integral $$J:=∫_-pi/2^0fracdxtan^s x + 1 $$
can have an imaginary part. There could be one answer for each branch of $x^s=e^slog x $. Since $x<0$, these branches are
$$ e^x = |x|^s e^(2k+1)À i s,quad k ∈ mathbb Z$$
giving
$$ frac1tan^s x + 1 = frac1e^(2k+1)Ài s tan^s(-x) + 1$$
If you try to go through the same proof, you end up with
$$ J = ∫_0^1 fracv^s(v^s + e^(2k+1)iÀ s)(v^2+1) + frac1(e^(2k+1)iÀ sv^s + 1)(v^2+1) dv$$
One can check that
$$ fracv^2s exp(ix) + 2v^s + exp(ix)[v^s+exp(ix)][v^s exp(ix)+1] = 1 -2 i sin(x) (v^2s + 2 v^s cos(x) + 1)$$
with $x = (2k+1)À i s$, this verifies that regardless of the branch cut used to define $tan^s x$ for $x<0$,
$$Re∫_-pi/2^0 frac1tan^s x + 1 dx = fracÀ4 $$
answered Jul 31 at 0:23
Calvin Khor
8,00411132
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
There will be no problem if you write $(tan^2 x)^n$. But you can try to insist on using the power $s:=2n$. If you were talking about $$2∫_0^pi/2fracdxtan^s x + 1$$ there would be no issue using your proof still, but the integral $$J:=∫_-pi/2^0fracdxtan^s x + 1 $$
can have an imaginary part. There could be one answer for each branch of $x^s=e^slog x $. Since $x<0$, these branches are
$$ e^x = |x|^s e^(2k+1)À i s,quad k ∈ mathbb Z$$
giving
$$ frac1tan^s x + 1 = frac1e^(2k+1)Ài s tan^s(-x) + 1$$
If you try to go through the same proof, you end up with
$$ J = ∫_0^1 fracv^s(v^s + e^(2k+1)iÀ s)(v^2+1) + frac1(e^(2k+1)iÀ sv^s + 1)(v^2+1) dv$$
One can check that
$$ fracv^2s exp(ix) + 2v^s + exp(ix)[v^s+exp(ix)][v^s exp(ix)+1] = 1 -2 i sin(x) (v^2s + 2 v^s cos(x) + 1)$$
with $x = (2k+1)À i s$, this verifies that regardless of the branch cut used to define $tan^s x$ for $x<0$,
$$Re∫_-pi/2^0 frac1tan^s x + 1 dx = fracÀ4 $$
answered Jul 31 at 0:23
Calvin Khor
8,00411132
There will be no problem if you write $(tan^2 x)^n$. But you can try to insist on using the power $s:=2n$. If you were talking about $$2∫_0^pi/2fracdxtan^s x + 1$$ there would be no issue using your proof still, but the integral $$J:=∫_-pi/2^0fracdxtan^s x + 1 $$
can have an imaginary part. There could be one answer for each branch of $x^s=e^slog x $. Since $x<0$, these branches are
$$ e^x = |x|^s e^(2k+1)À i s,quad k ∈ mathbb Z$$
giving
$$ frac1tan^s x + 1 = frac1e^(2k+1)Ài s tan^s(-x) + 1$$
If you try to go through the same proof, you end up with
$$ J = ∫_0^1 fracv^s(v^s + e^(2k+1)iÀ s)(v^2+1) + frac1(e^(2k+1)iÀ sv^s + 1)(v^2+1) dv$$
One can check that
$$ fracv^2s exp(ix) + 2v^s + exp(ix)[v^s+exp(ix)][v^s exp(ix)+1] = 1 -2 i sin(x) (v^2s + 2 v^s cos(x) + 1)$$
with $x = (2k+1)À i s$, this verifies that regardless of the branch cut used to define $tan^s x$ for $x<0$,
$$Re∫_-pi/2^0 frac1tan^s x + 1 dx = fracÀ4 $$
answered Jul 31 at 0:23
Calvin Khor
8,00411132
edited Jul 31 at 1:52
answered Jul 31 at 0:23
Calvin Khor
8,00411132
answered Jul 31 at 0:23
Calvin Khor
8,00411132
answered Jul 31 at 0:23
Calvin Khor
8,00411132
8,00411132
add a comment |Â
add a comment |Â
up vote
0
down vote
Your integral always equals
$$2int^infty_0frac1(1+u^2n)(1+u^2)du$$
$$=2int^1_0 frac1(1+u^2n)(1+u^2)du +2int^infty_1 frac1(1+u^2n)(1+u^2)du$$
By substitution $umapsto frac1u$ on the second integral, one obtains
$$2int^1_0fracfrac1u^2du(1+1/u^2n)(1+1/u^2)=2int^1_0fracu^2n(u^2n+1)(1+u^2)du$$
Adding two integrals together, you see the cancellation and obtain
$$2int^1_0frac11+u^2du=fracpi2$$
So, yes, your method does work for any $n$.
answered Jul 30 at 23:22
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
Your integral always equals
$$2int^infty_0frac1(1+u^2n)(1+u^2)du$$
$$=2int^1_0 frac1(1+u^2n)(1+u^2)du +2int^infty_1 frac1(1+u^2n)(1+u^2)du$$
By substitution $umapsto frac1u$ on the second integral, one obtains
$$2int^1_0fracfrac1u^2du(1+1/u^2n)(1+1/u^2)=2int^1_0fracu^2n(u^2n+1)(1+u^2)du$$
Adding two integrals together, you see the cancellation and obtain
$$2int^1_0frac11+u^2du=fracpi2$$
So, yes, your method does work for any $n$.
answered Jul 30 at 23:22
Szeto
3,8431421
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Your integral always equals
$$2int^infty_0frac1(1+u^2n)(1+u^2)du$$
$$=2int^1_0 frac1(1+u^2n)(1+u^2)du +2int^infty_1 frac1(1+u^2n)(1+u^2)du$$
By substitution $umapsto frac1u$ on the second integral, one obtains
$$2int^1_0fracfrac1u^2du(1+1/u^2n)(1+1/u^2)=2int^1_0fracu^2n(u^2n+1)(1+u^2)du$$
Adding two integrals together, you see the cancellation and obtain
$$2int^1_0frac11+u^2du=fracpi2$$
So, yes, your method does work for any $n$.
answered Jul 30 at 23:22
Szeto
3,8431421
Your integral always equals
$$2int^infty_0frac1(1+u^2n)(1+u^2)du$$
$$=2int^1_0 frac1(1+u^2n)(1+u^2)du +2int^infty_1 frac1(1+u^2n)(1+u^2)du$$
By substitution $umapsto frac1u$ on the second integral, one obtains
$$2int^1_0fracfrac1u^2du(1+1/u^2n)(1+1/u^2)=2int^1_0fracu^2n(u^2n+1)(1+u^2)du$$
Adding two integrals together, you see the cancellation and obtain
$$2int^1_0frac11+u^2du=fracpi2$$
So, yes, your method does work for any $n$.
answered Jul 30 at 23:22
Szeto
3,8431421
answered Jul 30 at 23:22
Szeto
3,8431421
answered Jul 30 at 23:22
Szeto
3,8431421
answered Jul 30 at 23:22
Szeto
3,8431421
3,8431421
add a comment |Â
add a comment |Â
related : Integrate $int_0^pi/2 frac11+tan^alphax,mathrmdx$
– Calvin Khor
Jul 30 at 23:23
Why is Wolfram giving an imaginary part? That's why i'm asking this question. Thanks for the attention.
– otreblig
Jul 31 at 0:12
If $ninmathbbR$, then the value of the integral is purely real.
– Mark Viola
Jul 31 at 0:13
@MarkViola, I'm not certain of my work in my answer below and OP could have been clearer, but I don't think OP is asking the same question, given the above comment. (edit: actually I just saw a mistake...)
– Calvin Khor
Jul 31 at 0:49
@otreblig I have verified your conjecture below
– Calvin Khor
Jul 31 at 1:53