Mellin transform of sin(z) and cos(z)
by math_explorer, Feb 28, 2014, 1:58 AM
This is the first nontrivial contour integral I did. You should know when they bring out the \mathcals that things are about to get serious. However, I procrastinated since I was nerd-sniped into doing codyj's CMC Problem 10, which burned through my mental capacity for LaTeXing integrals. The result of all that is here: https://brilliant.org/discussions/thread/cmc-problem-10/?ref_id=46730#comment-a3f525721e92 (Finally, they had the sanity to allow people to view discussions without an account.)
As you can see, that was a long time ago. Then a lot of other stuff happened and I forgot about this post. Until now as I'm scrambling to keep my February 2014 archive nonempty.
Oops.
(Note: I'm using
and 
to denote the real and imaginary parts of stuff because I don't have strong preferences for other notation and they're easiest to type --- \Re, \Im. I'm lazy.)
Anyway, it's pretty clear that
is more reasonable to deal with in its exponential form 
, and likewise for 
. So if we plug those in, we realize we want to calculate ![\[\int_0^\infty e^{it}t^{z-1} dt\]](//latex.artofproblemsolving.com/2/b/9/2b96fd8254382d26bbd65cec9ee461802926e8ec.png)
and its conjugate partner ![\[\int_0^\infty e^{-it}t^{z-1} dt.\]](//latex.artofproblemsolving.com/9/5/c/95cc7a07a71c7a4986889634c7c6f7561ccb33b4.png)
In fact, it's necessary since these are expressible in terms of and too.
The book is very nice and gives you the exact function and contour you want to use: integrate
across this.
![[asy]
pen rp = red + linewidth(1);
pen crim = rgb(0.6, 0, 0);
draw((-0.1,0)--(2.3,0), arrow=ArcArrow(SimpleHead));
draw((0,-0.1)--(0,2.3), arrow=ArcArrow(SimpleHead));
draw((0.4,0)--(2,0), rp, arrow=MidArcArrow(HookHead), L=Label("$\alpha$", position=MidPoint, align=2N, p=crim));
draw(arc((0,0), r=2, angle1=0, angle2=90), rp, arrow=MidArcArrow(HookHead), L=Label("$\beta$", position=MidPoint, align=2NE, p=crim));
draw((0,2)--(0,0.4), rp, arrow=MidArcArrow(HookHead), L=Label("$\gamma$", position=MidPoint, align=2E, p=crim));
draw(arc((0,0), r=0.4, angle1=90, angle2=0), rp, arrow=MidArcArrow(HookHead), L=Label("$\delta$", position=MidPoint, align=2NE, p=crim));
label("$\epsilon$", (0.4,0), align=S);
label("$R$", (2,0), align=S);
label("$0$", (0,0), align=SW);
[/asy]](//latex.artofproblemsolving.com/c/7/5/c75ebeeb755cac1956ccf2a986a3ac581a3b008d.png)
2014/11/20: I learned Asymptote! Here's the old image for teh lulz.
I've labeled the four parts of the path with greek letters. The integral across
parametrized as 
gives us ![\[ \int_R^\epsilon e^{-it}(it)^{z-1} i\,dt = -i^z \int_\epsilon^R e^{-it}t^{z-1} dt \]](//latex.artofproblemsolving.com/d/4/0/d407b78b2618bddf7e8616d6c3bd5a81d27fd822.png)
(is it normal that I can't do any variable substitutions in integrals, no matter how trivial, without writing this out every time?)
This obviously results in our target integral if we take
and 
, so we try taking those limits. Conveniently, the integral over 
simply goes to 
, so all we need to do is show that the integrals over 
and 
go to zero to get what the problem wants us to show (after some uninteresting algebraic manipulations). It turns out that this is quite nontrivial, however. For either curve, if 
is close to purely imaginary, then 
will have magnitude 1 and we have no idea how 
would behave (the growth of its magnitude depends on 
, which is not restricted).
Sidebar on exponents
We must be very careful with exponents here. The simplest definition of the exponent for complex numbers uses![\[ a^b = e^{b \log a}, \]](//latex.artofproblemsolving.com/1/7/7/177aacd94fe7e5b25456df204034135108db9150.png)
but 
is not generally well-defined for 
. It's fine if 
is positive and real, but we can't define 
continuously on the entire complex plane minus 
, because if you circle once you find that you end up with 
. So we usually take the principal branch, saying that the log of nonpositive numbers is undefined and that the imaginary part of 
for other numbers must always between 
and 
. There are (uncountably infinitely) other ways; as long as you cut a slit in the complex plane that prevents you from circling the origin, you're fine.
The upshot of all this for us:
. If is close to purely imaginary with an unknown magnitude, 
is only going to have a 
imaginary part, but when that imaginary part gets multiplied by 
, which has an unknown imaginary part, and the whole thing gets integrated over a really big circle, bad things might happen. We have to prove they won't.
-- end --
Let's deal with first. Let the radius be 
, and parametrize the arc in the obvious way with an angle:
We don't actually know if just taking the absolute value of the integrand like this won't make our inequality too weak, but there aren't obvious ways to tighten it, so let's try it. Troublingly,
goes to infinity if , so we have to get a strong-ish bound out of the integrand. We can't just bound 
by a constant --- that constant would have to be 0 and we'd have the unrestricted free to do whatever it wants again. So... we use the concavity of on the interval we're concerned with to bound it by a linear function.
![\[ \cos \theta \geq 1 - \frac{\theta}{\pi/2} \]](//latex.artofproblemsolving.com/1/7/a/17aedf518c273655a9442d814d9074476654b56b.png)
The result is something that we can actually integrate. Drum roll, please. Does itblend work...!?
The fraction on the left goes to zero because
grows faster than . The term in the parentheses to the right stays constant for large . So indeed, the integral over goes to zero.
For, it's really the same integral, just with a sign flip and with replaced by 
, which tends to zero. The right term still stays constant. The left fraction goes to zero if is nonzero...
Oh dear, if is zero then the fraction on the left will have a denominator that vanishes more strongly than its numerator as (which is actually ) tends to 0, which means the whole thing will diverge. That's not good.
However, luckily, if is zero then the term in parentheses to the right will also go to zero as aka tends to zero. This vanishing is order 1, as seen from differentiation, so that's enough to block the denominator on the left from destroying things.
...
Actually, the integral along can be seen to vanish just as easily from the original integral; even though 
diverges as tends to zero, it vanishes with order less than 1, so just the vanishing of the length of is enough to compensate. So we don't need to go through those contortions to get rid of . Darn why textbooks so troll
Anyway, after the uninteresting algebraic manipulations that I'm too lazy to type out, we're done.
As you can see, that was a long time ago. Then a lot of other stuff happened and I forgot about this post. Until now as I'm scrambling to keep my February 2014 archive nonempty.
Oops.
Quote:
10. An integral of the form ![\[ F(z) = \int_0^\infty f(t)t^{z-1} dt \]](//latex.artofproblemsolving.com/a/b/d/abd4b5656a4286ab99e46aa94e52905f7971d4fd.png)
is called a Mellin transform, and we shall write 
.
... Prove that![\[ \mathcal{M}(\cos)(z) = \int_0^\infty \cos(t)t^{z-1}dt = \Gamma(z)\cos\left(\pi\frac{z}{2}\right) \qquad \textrm{for } 0 < \Re(z) < 1 \]](//latex.artofproblemsolving.com/8/b/3/8b32452f3ff2178388b0b0ce8b57166bfc37e177.png)
and ![\[ \mathcal{M}(\sin)(z) = \int_0^\infty \sin(t)t^{z-1}dt = \Gamma(z)\sin\left(\pi\frac{z}{2}\right) \qquad \textrm{for } 0 < \Re(z) < 1. \]](//latex.artofproblemsolving.com/6/7/d/67dd940a3ca954b05d9bbc3a40e40cc870e913b5.png)
![\[ F(z) = \int_0^\infty f(t)t^{z-1} dt \]](http://latex.artofproblemsolving.com/a/b/d/abd4b5656a4286ab99e46aa94e52905f7971d4fd.png)
is called a Mellin transform, and we shall write 
.... Prove that
![\[ \mathcal{M}(\cos)(z) = \int_0^\infty \cos(t)t^{z-1}dt = \Gamma(z)\cos\left(\pi\frac{z}{2}\right) \qquad \textrm{for } 0 < \Re(z) < 1 \]](http://latex.artofproblemsolving.com/8/b/3/8b32452f3ff2178388b0b0ce8b57166bfc37e177.png)
and ![\[ \mathcal{M}(\sin)(z) = \int_0^\infty \sin(t)t^{z-1}dt = \Gamma(z)\sin\left(\pi\frac{z}{2}\right) \qquad \textrm{for } 0 < \Re(z) < 1. \]](http://latex.artofproblemsolving.com/6/7/d/67dd940a3ca954b05d9bbc3a40e40cc870e913b5.png)
(Note: I'm using
and 
to denote the real and imaginary parts of stuff because I don't have strong preferences for other notation and they're easiest to type --- \Re, \Im. I'm lazy.)Anyway, it's pretty clear that
is more reasonable to deal with in its exponential form 
, and likewise for 
. So if we plug those in, we realize we want to calculate ![\[\int_0^\infty e^{it}t^{z-1} dt\]](http://latex.artofproblemsolving.com/2/b/9/2b96fd8254382d26bbd65cec9ee461802926e8ec.png)
and its conjugate partner ![\[\int_0^\infty e^{-it}t^{z-1} dt.\]](http://latex.artofproblemsolving.com/9/5/c/95cc7a07a71c7a4986889634c7c6f7561ccb33b4.png)
In fact, it's necessary since these are expressible in terms of and too.The book is very nice and gives you the exact function and contour you want to use: integrate
across this.![[asy]
pen rp = red + linewidth(1);
pen crim = rgb(0.6, 0, 0);
draw((-0.1,0)--(2.3,0), arrow=ArcArrow(SimpleHead));
draw((0,-0.1)--(0,2.3), arrow=ArcArrow(SimpleHead));
draw((0.4,0)--(2,0), rp, arrow=MidArcArrow(HookHead), L=Label("$\alpha$", position=MidPoint, align=2N, p=crim));
draw(arc((0,0), r=2, angle1=0, angle2=90), rp, arrow=MidArcArrow(HookHead), L=Label("$\beta$", position=MidPoint, align=2NE, p=crim));
draw((0,2)--(0,0.4), rp, arrow=MidArcArrow(HookHead), L=Label("$\gamma$", position=MidPoint, align=2E, p=crim));
draw(arc((0,0), r=0.4, angle1=90, angle2=0), rp, arrow=MidArcArrow(HookHead), L=Label("$\delta$", position=MidPoint, align=2NE, p=crim));
label("$\epsilon$", (0.4,0), align=S);
label("$R$", (2,0), align=S);
label("$0$", (0,0), align=SW);
[/asy]](http://latex.artofproblemsolving.com/c/7/5/c75ebeeb755cac1956ccf2a986a3ac581a3b008d.png)
2014/11/20: I learned Asymptote! Here's the old image for teh lulz.
I've labeled the four parts of the path with greek letters. The integral across
parametrized as 
gives us ![\[ \int_R^\epsilon e^{-it}(it)^{z-1} i\,dt = -i^z \int_\epsilon^R e^{-it}t^{z-1} dt \]](http://latex.artofproblemsolving.com/d/4/0/d407b78b2618bddf7e8616d6c3bd5a81d27fd822.png)
(is it normal that I can't do any variable substitutions in integrals, no matter how trivial, without writing this out every time?)This obviously results in our target integral if we take
and 
, so we try taking those limits. Conveniently, the integral over 
simply goes to 
, so all we need to do is show that the integrals over 
and 
go to zero to get what the problem wants us to show (after some uninteresting algebraic manipulations). It turns out that this is quite nontrivial, however. For either curve, if 
is close to purely imaginary, then 
will have magnitude 1 and we have no idea how 
would behave (the growth of its magnitude depends on 
, which is not restricted).Sidebar on exponents
We must be very careful with exponents here. The simplest definition of the exponent for complex numbers uses
![\[ a^b = e^{b \log a}, \]](http://latex.artofproblemsolving.com/1/7/7/177aacd94fe7e5b25456df204034135108db9150.png)
but 
is not generally well-defined for 
. It's fine if 
is positive and real, but we can't define 
continuously on the entire complex plane minus 
, because if you circle once you find that you end up with 
. So we usually take the principal branch, saying that the log of nonpositive numbers is undefined and that the imaginary part of 
for other numbers must always between 
and 
. There are (uncountably infinitely) other ways; as long as you cut a slit in the complex plane that prevents you from circling the origin, you're fine.The upshot of all this for us:
. If is close to purely imaginary with an unknown magnitude, 
is only going to have a 
imaginary part, but when that imaginary part gets multiplied by 
, which has an unknown imaginary part, and the whole thing gets integrated over a really big circle, bad things might happen. We have to prove they won't.-- end --
Let's deal with
first. Let the radius be 
, and parametrize the arc in the obvious way with an angle:
We don't actually know if just taking the absolute value of the integrand like this won't make our inequality too weak, but there aren't obvious ways to tighten it, so let's try it. Troublingly,
goes to infinity if , so we have to get a strong-ish bound out of the integrand. We can't just bound 
by a constant --- that constant would have to be 0 and we'd have the unrestricted free to do whatever it wants again. So... we use the concavity of on the interval we're concerned with to bound it by a linear function.![\[ \cos \theta \geq 1 - \frac{\theta}{\pi/2} \]](http://latex.artofproblemsolving.com/1/7/a/17aedf518c273655a9442d814d9074476654b56b.png)
The result is something that we can actually integrate. Drum roll, please. Does it
The fraction on the left goes to zero because
grows faster than . The term in the parentheses to the right stays constant for large . So indeed, the integral over goes to zero.For
, it's really the same integral, just with a sign flip and with replaced by 
, which tends to zero. The right term still stays constant. The left fraction goes to zero if is nonzero...Oh dear, if
is zero then the fraction on the left will have a denominator that vanishes more strongly than its numerator as (which is actually ) tends to 0, which means the whole thing will diverge. That's not good.However, luckily, if
is zero then the term in parentheses to the right will also go to zero as aka tends to zero. This vanishing is order 1, as seen from differentiation, so that's enough to block the denominator on the left from destroying things....
Actually, the integral along
can be seen to vanish just as easily from the original integral; even though 
diverges as tends to zero, it vanishes with order less than 1, so just the vanishing of the length of is enough to compensate. So we don't need to go through those contortions to get rid of . Darn why textbooks so trollAnyway, after the uninteresting algebraic manipulations that I'm too lazy to type out, we're done.
This post has been edited 3 times. Last edited by math_explorer, Apr 26, 2015, 10:47 AM
May 2020
January 2020