Mixing
Bringing infinite limit inside integral
$begingroup$
I want to find $$lim_{R to infty} int_{0}^{pi} e^{-Rsin (t)}dt.$$ Since $sin(t)$ is nonnegative on $[0,pi]$ the integrand vanishes as $R rightarrow infty$. So I want to bring the limit under the integral to conclude that the limit is zero, but I'm not sure if it's justified here. I'm not particularly well-versed in this kind of problem so I would appreciate any relevant results/theorems that would help here too. Thanks!
real-analysis integration
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
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add a comment |
$begingroup$
I want to find $$lim_{R to infty} int_{0}^{pi} e^{-Rsin (t)}dt.$$ Since $sin(t)$ is nonnegative on $[0,pi]$ the integrand vanishes as $R rightarrow infty$. So I want to bring the limit under the integral to conclude that the limit is zero, but I'm not sure if it's justified here. I'm not particularly well-versed in this kind of problem so I would appreciate any relevant results/theorems that would help here too. Thanks!
real-analysis integration
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
$endgroup$
1
$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
1
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16
add a comment |
$begingroup$
I want to find $$lim_{R to infty} int_{0}^{pi} e^{-Rsin (t)}dt.$$ Since $sin(t)$ is nonnegative on $[0,pi]$ the integrand vanishes as $R rightarrow infty$. So I want to bring the limit under the integral to conclude that the limit is zero, but I'm not sure if it's justified here. I'm not particularly well-versed in this kind of problem so I would appreciate any relevant results/theorems that would help here too. Thanks!
real-analysis integration
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
$endgroup$
I want to find $$lim_{R to infty} int_{0}^{pi} e^{-Rsin (t)}dt.$$ Since $sin(t)$ is nonnegative on $[0,pi]$ the integrand vanishes as $R rightarrow infty$. So I want to bring the limit under the integral to conclude that the limit is zero, but I'm not sure if it's justified here. I'm not particularly well-versed in this kind of problem so I would appreciate any relevant results/theorems that would help here too. Thanks!
real-analysis integration
real-analysis integration
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
asked Jan 29 at 21:53
Alex SangerAlex Sanger
10329
10329
1
$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
1
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16
add a comment |
1
$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
1
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16
1
1
$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
1
1
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $sin(x) geq frac{2}{pi} x $ for $x in [0,pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ int limits_0^pi mathrm{e}^{-R sin(x)} , mathrm{d} x = 2 int limits_0^{pi/2} mathrm{e}^{-R sin(x)} , mathrm{d} x leq 2 int limits_0^{pi/2} mathrm{e}^{-R frac{2}{pi} x} , mathrm{d} x , .$$
The latter integral can be computed explicitly and will lead to the desired limit.
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
$endgroup$
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
add a comment |
$begingroup$
To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $sin$ is zero, so let's fold it over with symmetry:
$$I(R) = int_0^{pi} e^{-Rsin t},dt = 2int_0^{pi/2}e^{-Rsin t},dt$$
Now, we'll split that $int_0^{pi/2}$ into $int_0^{epsilon/4}+int_{epsilon/4}^{pi/2}$:
begin{align*}I(R) &= 2int_0^{epsilon/4}e^{-Rsin t},dt+2int_{epsilon/4}^{pi/2}e^{-Rsin t},dt\
I(R) &le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt = frac{epsilon}{2}+left(pi-frac{epsilon}{2}right)expleft(-Rsin frac{epsilon}{4}right)end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $sinfrac{epsilon}{4}approx frac{epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $sin$ by the secant line: $sin x ge frac{2}{pi}x$ for $0le xle frac{pi}{2}$. This leads to $expleft(-Rsin frac{epsilon}{4}right)le expleft(-Rcdotfrac{2}{pi}cdotfrac{epsilon}{4}right)$. Then of course $pi-frac{epsilon}{2}le pi$, and we can make this term less than $frac{epsilon}{2}$ simply by choosing $R$ large enough. For $Rge frac{2pi}{epsilon}ln(2pi)$,
$$I(R) le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt le frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$
As $epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $frac{epsilon}{4}$? Well, you'll probably start out by writing $epsilon$ - and then when you get an estimate of $4epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
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add a comment |
$begingroup$
Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n to f$ uniformly on $[a,b]$, then
$int_{a}^{b} f_n to int_{a}^{b} f$.
To show that
$$
lim_{R to infty} int_{0}^{pi} e^{-R sin(t)} , dt = 0,
$$
we will show that for all $varepsilon > 0$, there exists $R_0$ so that if $R geq R_0$, then
$$
int_{0}^{pi} e^{-R sin(t)} , dt leq varepsilon.
$$
Note: the family of functions ${ e^{-R sin(t)} }_{R > 0}$ converges uniformly to $0$ on the interval $[varepsilon/4, pi - varepsilon/4]$, so
$$
lim_{R to infty} int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} ,dt = 0.
$$
That is, there is some $R_0$ so that if $R geq R_0$, then
$$
int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt leq frac{varepsilon}{2}.
$$
For $R geq R_0$, we have
$$
int_{0}^{pi} e^{-R sin(t)} , dt = int_0^{varepsilon/4} e^{-R sin(t)} , dt
+ int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt leq int_0^{varepsilon/4} 1 , dt + frac{varepsilon}{2} + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt= varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, varepsilon/4]$ and $[pi - varepsilon/4, pi]$.)
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
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Ah, you’re right. I’ve edited.
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– Jordan Green
Jan 29 at 22:45
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
lim_{R to infty}int_{0}^{pi}expo{-Rsinpars{t}},dd t & =
lim_{R to infty}int_{-pi/2}^{pi/2}expo{-Rcospars{t}},dd t =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rcospars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rsinpars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{infty}expo{-Rt},dd t
qquadpars{~Laplace's Method~}
\[5mm] & =
2lim_{R to infty}{1 over R} = bbx{0}
end{align}
Laplace's Method.
Indeed,
$$
int_{0}^{pi}expo{-Rsinpars{t}},dd t =
{1 over 2},pibracks{mrm{I}_{0}pars{R} - mrm{L}_{0}pars{R}}
$$
where $ds{mrm{I}_{nu}}$ and $ds{mrm{L}_{nu}}$ are Bessel and Struve Functions, respectively.
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
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add a comment |
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4 Answers
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active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $sin(x) geq frac{2}{pi} x $ for $x in [0,pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ int limits_0^pi mathrm{e}^{-R sin(x)} , mathrm{d} x = 2 int limits_0^{pi/2} mathrm{e}^{-R sin(x)} , mathrm{d} x leq 2 int limits_0^{pi/2} mathrm{e}^{-R frac{2}{pi} x} , mathrm{d} x , .$$
The latter integral can be computed explicitly and will lead to the desired limit.
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
$endgroup$
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
add a comment |
$begingroup$
As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $sin(x) geq frac{2}{pi} x $ for $x in [0,pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ int limits_0^pi mathrm{e}^{-R sin(x)} , mathrm{d} x = 2 int limits_0^{pi/2} mathrm{e}^{-R sin(x)} , mathrm{d} x leq 2 int limits_0^{pi/2} mathrm{e}^{-R frac{2}{pi} x} , mathrm{d} x , .$$
The latter integral can be computed explicitly and will lead to the desired limit.
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
$endgroup$
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
add a comment |
$begingroup$
As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $sin(x) geq frac{2}{pi} x $ for $x in [0,pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ int limits_0^pi mathrm{e}^{-R sin(x)} , mathrm{d} x = 2 int limits_0^{pi/2} mathrm{e}^{-R sin(x)} , mathrm{d} x leq 2 int limits_0^{pi/2} mathrm{e}^{-R frac{2}{pi} x} , mathrm{d} x , .$$
The latter integral can be computed explicitly and will lead to the desired limit.
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
$endgroup$
As mentioned in the comments, the dominated convergence theorem is probably the fastest way to justify the interchange of limit and integral. In this particular case that is not necessary, however, as there is an easier method:
You can use the inequality $sin(x) geq frac{2}{pi} x $ for $x in [0,pi/2]$ (which holds since the sine function is concave on this interval) to find
$$ int limits_0^pi mathrm{e}^{-R sin(x)} , mathrm{d} x = 2 int limits_0^{pi/2} mathrm{e}^{-R sin(x)} , mathrm{d} x leq 2 int limits_0^{pi/2} mathrm{e}^{-R frac{2}{pi} x} , mathrm{d} x , .$$
The latter integral can be computed explicitly and will lead to the desired limit.
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
answered Jan 29 at 22:23
ComplexYetTrivialComplexYetTrivial
5,0082631
5,0082631
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
add a comment |
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
$begingroup$
I wouldn’t have thought to write the integral over $[0, pi]$ as twice the integral over $[0, pi/2]$. This was very clever.
$endgroup$
– Jordan Green
Jan 29 at 22:38
add a comment |
$begingroup$
To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $sin$ is zero, so let's fold it over with symmetry:
$$I(R) = int_0^{pi} e^{-Rsin t},dt = 2int_0^{pi/2}e^{-Rsin t},dt$$
Now, we'll split that $int_0^{pi/2}$ into $int_0^{epsilon/4}+int_{epsilon/4}^{pi/2}$:
begin{align*}I(R) &= 2int_0^{epsilon/4}e^{-Rsin t},dt+2int_{epsilon/4}^{pi/2}e^{-Rsin t},dt\
I(R) &le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt = frac{epsilon}{2}+left(pi-frac{epsilon}{2}right)expleft(-Rsin frac{epsilon}{4}right)end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $sinfrac{epsilon}{4}approx frac{epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $sin$ by the secant line: $sin x ge frac{2}{pi}x$ for $0le xle frac{pi}{2}$. This leads to $expleft(-Rsin frac{epsilon}{4}right)le expleft(-Rcdotfrac{2}{pi}cdotfrac{epsilon}{4}right)$. Then of course $pi-frac{epsilon}{2}le pi$, and we can make this term less than $frac{epsilon}{2}$ simply by choosing $R$ large enough. For $Rge frac{2pi}{epsilon}ln(2pi)$,
$$I(R) le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt le frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$
As $epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $frac{epsilon}{4}$? Well, you'll probably start out by writing $epsilon$ - and then when you get an estimate of $4epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $sin$ is zero, so let's fold it over with symmetry:
$$I(R) = int_0^{pi} e^{-Rsin t},dt = 2int_0^{pi/2}e^{-Rsin t},dt$$
Now, we'll split that $int_0^{pi/2}$ into $int_0^{epsilon/4}+int_{epsilon/4}^{pi/2}$:
begin{align*}I(R) &= 2int_0^{epsilon/4}e^{-Rsin t},dt+2int_{epsilon/4}^{pi/2}e^{-Rsin t},dt\
I(R) &le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt = frac{epsilon}{2}+left(pi-frac{epsilon}{2}right)expleft(-Rsin frac{epsilon}{4}right)end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $sinfrac{epsilon}{4}approx frac{epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $sin$ by the secant line: $sin x ge frac{2}{pi}x$ for $0le xle frac{pi}{2}$. This leads to $expleft(-Rsin frac{epsilon}{4}right)le expleft(-Rcdotfrac{2}{pi}cdotfrac{epsilon}{4}right)$. Then of course $pi-frac{epsilon}{2}le pi$, and we can make this term less than $frac{epsilon}{2}$ simply by choosing $R$ large enough. For $Rge frac{2pi}{epsilon}ln(2pi)$,
$$I(R) le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt le frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$
As $epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $frac{epsilon}{4}$? Well, you'll probably start out by writing $epsilon$ - and then when you get an estimate of $4epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
$endgroup$
add a comment |
$begingroup$
To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $sin$ is zero, so let's fold it over with symmetry:
$$I(R) = int_0^{pi} e^{-Rsin t},dt = 2int_0^{pi/2}e^{-Rsin t},dt$$
Now, we'll split that $int_0^{pi/2}$ into $int_0^{epsilon/4}+int_{epsilon/4}^{pi/2}$:
begin{align*}I(R) &= 2int_0^{epsilon/4}e^{-Rsin t},dt+2int_{epsilon/4}^{pi/2}e^{-Rsin t},dt\
I(R) &le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt = frac{epsilon}{2}+left(pi-frac{epsilon}{2}right)expleft(-Rsin frac{epsilon}{4}right)end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $sinfrac{epsilon}{4}approx frac{epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $sin$ by the secant line: $sin x ge frac{2}{pi}x$ for $0le xle frac{pi}{2}$. This leads to $expleft(-Rsin frac{epsilon}{4}right)le expleft(-Rcdotfrac{2}{pi}cdotfrac{epsilon}{4}right)$. Then of course $pi-frac{epsilon}{2}le pi$, and we can make this term less than $frac{epsilon}{2}$ simply by choosing $R$ large enough. For $Rge frac{2pi}{epsilon}ln(2pi)$,
$$I(R) le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt le frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$
As $epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $frac{epsilon}{4}$? Well, you'll probably start out by writing $epsilon$ - and then when you get an estimate of $4epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
$endgroup$
To find this limit directly, we divide and conquer with a multi-part estimate; we split the interval into a narrow piece where the integral is small because the interval is small, and a wide piece where the integral is small because the function is small.
First off, I'd rather not deal with both endpoints where $sin$ is zero, so let's fold it over with symmetry:
$$I(R) = int_0^{pi} e^{-Rsin t},dt = 2int_0^{pi/2}e^{-Rsin t},dt$$
Now, we'll split that $int_0^{pi/2}$ into $int_0^{epsilon/4}+int_{epsilon/4}^{pi/2}$:
begin{align*}I(R) &= 2int_0^{epsilon/4}e^{-Rsin t},dt+2int_{epsilon/4}^{pi/2}e^{-Rsin t},dt\
I(R) &le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt = frac{epsilon}{2}+left(pi-frac{epsilon}{2}right)expleft(-Rsin frac{epsilon}{4}right)end{align*}
In both cases, we estimated the function by simply saying that it's less than the largest value on that subinterval.
So now, we need to estimate that exponential term. We have $sinfrac{epsilon}{4}approx frac{epsilon}{4}$ - but if we turn that into an inequality, it's a $<$, and that points in the wrong direction for what we need. So instead, estimate $sin$ by the secant line: $sin x ge frac{2}{pi}x$ for $0le xle frac{pi}{2}$. This leads to $expleft(-Rsin frac{epsilon}{4}right)le expleft(-Rcdotfrac{2}{pi}cdotfrac{epsilon}{4}right)$. Then of course $pi-frac{epsilon}{2}le pi$, and we can make this term less than $frac{epsilon}{2}$ simply by choosing $R$ large enough. For $Rge frac{2pi}{epsilon}ln(2pi)$,
$$I(R) le 2int_0^{epsilon/4}1,dt + 2int_{epsilon/4}^{pi/2}expleft(-Rsin frac{epsilon}{4}right),dt le frac{epsilon}{2}+frac{epsilon}{2}=epsilon$$
As $epsilon>0$ was arbitrary, that's the definition of convergence to zero, and we're done.
The dominated convergence theorem is of course the easy way; we just use the pointwise convergence to $0$ on the interior (and $1$ at the endpoints) and the fact that all of the functions are less than the integrable function $1$. Then again, this style of argument is worth knowing - it's how we prove the dominated convergence theorem, after all. Running through a few examples like this isn't a bad idea.
Why $frac{epsilon}{4}$? Well, you'll probably start out by writing $epsilon$ - and then when you get an estimate of $4epsilon$ for the whole thing, go back and divide them all by $4$. The final form of the argument doesn't have to match the scratch version.
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
answered Jan 29 at 22:40
jmerryjmerry
16.9k11633
16.9k11633
add a comment |
add a comment |
$begingroup$
Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n to f$ uniformly on $[a,b]$, then
$int_{a}^{b} f_n to int_{a}^{b} f$.
To show that
$$
lim_{R to infty} int_{0}^{pi} e^{-R sin(t)} , dt = 0,
$$
we will show that for all $varepsilon > 0$, there exists $R_0$ so that if $R geq R_0$, then
$$
int_{0}^{pi} e^{-R sin(t)} , dt leq varepsilon.
$$
Note: the family of functions ${ e^{-R sin(t)} }_{R > 0}$ converges uniformly to $0$ on the interval $[varepsilon/4, pi - varepsilon/4]$, so
$$
lim_{R to infty} int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} ,dt = 0.
$$
That is, there is some $R_0$ so that if $R geq R_0$, then
$$
int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt leq frac{varepsilon}{2}.
$$
For $R geq R_0$, we have
$$
int_{0}^{pi} e^{-R sin(t)} , dt = int_0^{varepsilon/4} e^{-R sin(t)} , dt
+ int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt leq int_0^{varepsilon/4} 1 , dt + frac{varepsilon}{2} + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt= varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, varepsilon/4]$ and $[pi - varepsilon/4, pi]$.)
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
add a comment |
$begingroup$
Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n to f$ uniformly on $[a,b]$, then
$int_{a}^{b} f_n to int_{a}^{b} f$.
To show that
$$
lim_{R to infty} int_{0}^{pi} e^{-R sin(t)} , dt = 0,
$$
we will show that for all $varepsilon > 0$, there exists $R_0$ so that if $R geq R_0$, then
$$
int_{0}^{pi} e^{-R sin(t)} , dt leq varepsilon.
$$
Note: the family of functions ${ e^{-R sin(t)} }_{R > 0}$ converges uniformly to $0$ on the interval $[varepsilon/4, pi - varepsilon/4]$, so
$$
lim_{R to infty} int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} ,dt = 0.
$$
That is, there is some $R_0$ so that if $R geq R_0$, then
$$
int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt leq frac{varepsilon}{2}.
$$
For $R geq R_0$, we have
$$
int_{0}^{pi} e^{-R sin(t)} , dt = int_0^{varepsilon/4} e^{-R sin(t)} , dt
+ int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt leq int_0^{varepsilon/4} 1 , dt + frac{varepsilon}{2} + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt= varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, varepsilon/4]$ and $[pi - varepsilon/4, pi]$.)
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
$endgroup$
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
add a comment |
$begingroup$
Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n to f$ uniformly on $[a,b]$, then
$int_{a}^{b} f_n to int_{a}^{b} f$.
To show that
$$
lim_{R to infty} int_{0}^{pi} e^{-R sin(t)} , dt = 0,
$$
we will show that for all $varepsilon > 0$, there exists $R_0$ so that if $R geq R_0$, then
$$
int_{0}^{pi} e^{-R sin(t)} , dt leq varepsilon.
$$
Note: the family of functions ${ e^{-R sin(t)} }_{R > 0}$ converges uniformly to $0$ on the interval $[varepsilon/4, pi - varepsilon/4]$, so
$$
lim_{R to infty} int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} ,dt = 0.
$$
That is, there is some $R_0$ so that if $R geq R_0$, then
$$
int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt leq frac{varepsilon}{2}.
$$
For $R geq R_0$, we have
$$
int_{0}^{pi} e^{-R sin(t)} , dt = int_0^{varepsilon/4} e^{-R sin(t)} , dt
+ int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt leq int_0^{varepsilon/4} 1 , dt + frac{varepsilon}{2} + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt= varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, varepsilon/4]$ and $[pi - varepsilon/4, pi]$.)
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
$endgroup$
Here is an elementary solution that does not involve the dominated convergence theorem but instead uses the fact if $f_n to f$ uniformly on $[a,b]$, then
$int_{a}^{b} f_n to int_{a}^{b} f$.
To show that
$$
lim_{R to infty} int_{0}^{pi} e^{-R sin(t)} , dt = 0,
$$
we will show that for all $varepsilon > 0$, there exists $R_0$ so that if $R geq R_0$, then
$$
int_{0}^{pi} e^{-R sin(t)} , dt leq varepsilon.
$$
Note: the family of functions ${ e^{-R sin(t)} }_{R > 0}$ converges uniformly to $0$ on the interval $[varepsilon/4, pi - varepsilon/4]$, so
$$
lim_{R to infty} int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} ,dt = 0.
$$
That is, there is some $R_0$ so that if $R geq R_0$, then
$$
int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt leq frac{varepsilon}{2}.
$$
For $R geq R_0$, we have
$$
int_{0}^{pi} e^{-R sin(t)} , dt = int_0^{varepsilon/4} e^{-R sin(t)} , dt
+ int_{varepsilon/4}^{pi - varepsilon/4} e^{-R sin(t)} , dt + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt leq int_0^{varepsilon/4} 1 , dt + frac{varepsilon}{2} + int_{pi -varepsilon/4}^{pi} e^{-R sin(t)} , dt= varepsilon.
$$
(Here, we are using $1$ as an upper bound for the integrand on the intervals $[0, varepsilon/4]$ and $[pi - varepsilon/4, pi]$.)
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
edited Jan 29 at 22:47
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
answered Jan 29 at 22:30
Jordan GreenJordan Green
1,146410
1,146410
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
add a comment |
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Technically, it's a bit off; the uniform convergence doesn't work if we extend that interval all the way to $pi$.
$endgroup$
– jmerry
Jan 29 at 22:41
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
$begingroup$
Ah, you’re right. I’ve edited.
$endgroup$
– Jordan Green
Jan 29 at 22:45
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
lim_{R to infty}int_{0}^{pi}expo{-Rsinpars{t}},dd t & =
lim_{R to infty}int_{-pi/2}^{pi/2}expo{-Rcospars{t}},dd t =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rcospars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rsinpars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{infty}expo{-Rt},dd t
qquadpars{~Laplace's Method~}
\[5mm] & =
2lim_{R to infty}{1 over R} = bbx{0}
end{align}
Laplace's Method.
Indeed,
$$
int_{0}^{pi}expo{-Rsinpars{t}},dd t =
{1 over 2},pibracks{mrm{I}_{0}pars{R} - mrm{L}_{0}pars{R}}
$$
where $ds{mrm{I}_{nu}}$ and $ds{mrm{L}_{nu}}$ are Bessel and Struve Functions, respectively.
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
lim_{R to infty}int_{0}^{pi}expo{-Rsinpars{t}},dd t & =
lim_{R to infty}int_{-pi/2}^{pi/2}expo{-Rcospars{t}},dd t =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rcospars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rsinpars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{infty}expo{-Rt},dd t
qquadpars{~Laplace's Method~}
\[5mm] & =
2lim_{R to infty}{1 over R} = bbx{0}
end{align}
Laplace's Method.
Indeed,
$$
int_{0}^{pi}expo{-Rsinpars{t}},dd t =
{1 over 2},pibracks{mrm{I}_{0}pars{R} - mrm{L}_{0}pars{R}}
$$
where $ds{mrm{I}_{nu}}$ and $ds{mrm{L}_{nu}}$ are Bessel and Struve Functions, respectively.
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
lim_{R to infty}int_{0}^{pi}expo{-Rsinpars{t}},dd t & =
lim_{R to infty}int_{-pi/2}^{pi/2}expo{-Rcospars{t}},dd t =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rcospars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rsinpars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{infty}expo{-Rt},dd t
qquadpars{~Laplace's Method~}
\[5mm] & =
2lim_{R to infty}{1 over R} = bbx{0}
end{align}
Laplace's Method.
Indeed,
$$
int_{0}^{pi}expo{-Rsinpars{t}},dd t =
{1 over 2},pibracks{mrm{I}_{0}pars{R} - mrm{L}_{0}pars{R}}
$$
where $ds{mrm{I}_{nu}}$ and $ds{mrm{L}_{nu}}$ are Bessel and Struve Functions, respectively.
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
lim_{R to infty}int_{0}^{pi}expo{-Rsinpars{t}},dd t & =
lim_{R to infty}int_{-pi/2}^{pi/2}expo{-Rcospars{t}},dd t =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rcospars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{pi/2}expo{-Rsinpars{t}},dd t
\[5mm] & =
2lim_{R to infty}int_{0}^{infty}expo{-Rt},dd t
qquadpars{~Laplace's Method~}
\[5mm] & =
2lim_{R to infty}{1 over R} = bbx{0}
end{align}
Laplace's Method.
Indeed,
$$
int_{0}^{pi}expo{-Rsinpars{t}},dd t =
{1 over 2},pibracks{mrm{I}_{0}pars{R} - mrm{L}_{0}pars{R}}
$$
where $ds{mrm{I}_{nu}}$ and $ds{mrm{L}_{nu}}$ are Bessel and Struve Functions, respectively.
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
edited Jan 31 at 4:44
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
answered Jan 29 at 23:43
Felix MarinFelix Marin
68.8k7110146
68.8k7110146
add a comment |
add a comment |
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$begingroup$
Maybe here: math.stackexchange.com/questions/614941/…
$endgroup$
– Tito Eliatron
Jan 29 at 21:55
1
$begingroup$
Are you familiar with the Lebesgue dominated convergence theorem? That could be useful here.
$endgroup$
– Jordan Green
Jan 29 at 22:04
$begingroup$
@JordanGreen I am not. This integral came up in a course in complex analysis I am taking.
$endgroup$
– Alex Sanger
Jan 29 at 22:16