Tricky real integral: $int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi$
up vote
8
down vote
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I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$
Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.
How can I tackle this integral?
calculus integration definite-integrals
add a comment |
up vote
8
down vote
favorite
I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$
Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.
How can I tackle this integral?
calculus integration definite-integrals
2
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
1
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20
add a comment |
up vote
8
down vote
favorite
up vote
8
down vote
favorite
I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$
Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.
How can I tackle this integral?
calculus integration definite-integrals
I'm trying to prove the following:
$$ int_0^{2 pi} e^{cos(2 t)} cos(sin(2 t)) =2pi $$
Numerical analysis agrees with this to very high accuracy, so I'm almost sure it's true. Mathematica gives this answer after thinking for a long, but gives an insane antiderivative in terms of exponential integrals. I'd like to evaluate the integral with purely real methods (I've never done complex analysis), as elegantly as possible.
How can I tackle this integral?
calculus integration definite-integrals
calculus integration definite-integrals
edited Dec 1 at 9:34
Martin Sleziak
44.5k7115268
44.5k7115268
asked Dec 1 at 2:15
NMister
376110
376110
2
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
1
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20
add a comment |
2
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
1
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20
2
2
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
1
1
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20
add a comment |
3 Answers
3
active
oldest
votes
up vote
22
down vote
Write
$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$
Then $I(0) = 2pi$, and for $alpha > 0$,
begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}
So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.
A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by
$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$
Then
$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$
and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.
add a comment |
up vote
5
down vote
Assuming that you could enjoy special functions.
Consider
$$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$
$$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
$$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
$$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
$$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
t}right)right)$$
$$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.
add a comment |
up vote
0
down vote
Put $z=e^{it}$ and using the formulas:
$$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$
$$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$
apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
22
down vote
Write
$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$
Then $I(0) = 2pi$, and for $alpha > 0$,
begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}
So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.
A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by
$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$
Then
$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$
and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.
add a comment |
up vote
22
down vote
Write
$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$
Then $I(0) = 2pi$, and for $alpha > 0$,
begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}
So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.
A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by
$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$
Then
$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$
and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.
add a comment |
up vote
22
down vote
up vote
22
down vote
Write
$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$
Then $I(0) = 2pi$, and for $alpha > 0$,
begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}
So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.
A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by
$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$
Then
$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$
and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.
Write
$$ I(alpha) = int_{0}^{2pi} e^{alpha cos(2t)}cos(alpha sin(2t)) , dt. $$
Then $I(0) = 2pi$, and for $alpha > 0$,
begin{align*}
I'(alpha)
&= int_{0}^{2pi} left[ e^{alpha cos(2t)}cos(alpha sin(2t))cos(2t) - e^{alpha cos(2t)}sin(alpha sin(2t))sin(2t) right] , dt \
&= left[ frac{1}{2alpha} e^{alphacos(2t)}sin(alphasin(2t)) right]_{0}^{2pi} \
&= 0.
end{align*}
So $I(alpha) = 2pi$ for all $alpha in mathbb{R}$.
A general computation. Let $f$ be analytic on $B(0,R)$. Define $I : [0, R) to mathbb{C}$ by
$$ I(r) = int_{0}^{2pi} fleft(re^{itheta}right) , dtheta. $$
Then
$$ I'(r)
= int_{0}^{2pi} f'left(re^{itheta}right)e^{itheta} , dtheta
= left[ frac{1}{ir} fleft(re^{itheta}right) right]_{0}^{2pi}
= 0 $$
and thus $I$ is constant with the value $I(0) = 2pi f(0)$. The above answer corresponds to the real part of this computation with $f(z) = e^z$.
edited Dec 1 at 9:13
Mutantoe
546411
546411
answered Dec 1 at 2:42
Sangchul Lee
90.8k12163263
90.8k12163263
add a comment |
add a comment |
up vote
5
down vote
Assuming that you could enjoy special functions.
Consider
$$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$
$$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
$$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
$$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
$$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
t}right)right)$$
$$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.
add a comment |
up vote
5
down vote
Assuming that you could enjoy special functions.
Consider
$$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$
$$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
$$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
$$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
$$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
t}right)right)$$
$$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.
add a comment |
up vote
5
down vote
up vote
5
down vote
Assuming that you could enjoy special functions.
Consider
$$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$
$$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
$$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
$$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
$$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
t}right)right)$$
$$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.
Assuming that you could enjoy special functions.
Consider
$$I=int e^{cos(a t)} cos(sin(a t)),dtqquad text{and}qquad J=int e^{cos(a t)} sin(sin(a t)),dt$$
$$I+iJ=int e^{e^{ i at}},dt=-frac{i}{a}, text{Ei}left(e^{i a t}right)$$
$$I-iJ=int e^{e^{- ia t}},dt=frac{i}{a} , text{Ei}left(e^{- i a t}right)$$ (where appear the exponential integral function) since, using $e^{kt}=u$,
$$int e^{e^{kt}},dt=frac{1}{k }intfrac{e^u}{u},du=frac{1}{k },text{Ei}(u)$$ This makes
$$I=frac{i }{2 a},left(text{Ei}left(e^{-i a t}right)-text{Ei}left(e^{i a
t}right)right)$$
$$J=-frac{1}{2 a},left(text{Ei}left(e^{-i a t}right)+text{Ei}left(e^{i a t}right)right)$$ For integer values of $a$, the definite integration from $0$ to $2pi$ requires breaking it in $2a$ intervals and, as @user170231 commented, the result is $2pi$ for any $a$.
edited Dec 1 at 15:21
answered Dec 1 at 5:35
Claude Leibovici
117k1156131
117k1156131
add a comment |
add a comment |
up vote
0
down vote
Put $z=e^{it}$ and using the formulas:
$$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$
$$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$
apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$
add a comment |
up vote
0
down vote
Put $z=e^{it}$ and using the formulas:
$$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$
$$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$
apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$
add a comment |
up vote
0
down vote
up vote
0
down vote
Put $z=e^{it}$ and using the formulas:
$$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$
$$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$
apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$
Put $z=e^{it}$ and using the formulas:
$$cos{2t}=frac{z^2+frac{1}{z^2}}{2}$$
$$sin{2t}=frac{z^2-frac{1}{z^2}}{2i}$$
apply the $text{Residue theorem}$ integrating along the circle $C(0,1)={z:|z|=1}$
answered Dec 1 at 15:38
Marios Gretsas
8,42511437
8,42511437
add a comment |
add a comment |
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2
Have you heard of Cauchy integral theorem?
– Frank W.
Dec 1 at 2:35
It would appear that$$int_0^{2pi}e^{cos at}cos(sin at),mathrm dt=2pi$$at least for all non-zero integer values of $a$.
– user170231
Dec 1 at 2:41
$x=cos(2t)$ then $dx=-2sin(2t)dt$ and the integral has to be broken up into four parts at intervals with $t=frac{pi}{2}$, each with integrand $frac{1}{2}e^xcos(x)$. Each piece should integrate to $frac{pi}{2}$.
– herb steinberg
Dec 1 at 2:48
1
You may not have done "complex analysis", in the sense of things like the residue theorem, but I can offer you a one-line proof that in terms of complex numbers only uses $exp ix=cos x+isin x$, viz. $$Reint_0^{2pi}exp(exp i2t)dt=Resum_{nge 0}frac{1}{n!}int_0^{2pi}exp i2nt dt=Resum_{nge 0}frac{2pidelta_{2n,,0}}{n!}=2pi.$$With a bit of care, you can use Taylor series to rewrite that as a real-only proof.
– J.G.
Dec 1 at 9:20