Putnam Problem of the Day

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10589427]

[math]
\text{Show that the improper integral}
\\
\quad \qquad \qquad \boxed{\lim_{B\to\infty}\int_{0}^B \sin(x) \sin(x^2)\,dx}
\\
\text{converges.}
[/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10589428]

Previous Thread >>10583271

[Anonymous No.10589752]

2hard

[Anonymous No.10589893]

>>10589427
Use the product of sines is a diference of cosines formula. Do a u substituition to turn it into a difference of Fresnel Integrals. By the convergence of fresnel integrals, this integral converges conditionally.

You can probably break the integral into a sum of integrals such that they form an alternating sequence going to zero (aka, apply the same trick for the usual proof of the convergence of the fresnel integrals) or a sequence you can compare to another integral. It's a bit tricky because of the beats like effect that comes from the product, but it's probably doable

[Anonymous No.10589976]

https://en.m.wikipedia.org/wiki/Fresnel_integral
As far as I got

[Anonymous No.10590002]

>>10589893
Do you even need to do any substitution?
-sin(x^2) < sin(x)sin(x^2) < sin(x^2). The integrals of the bounds are finite, so we are done.

[Anonymous No.10590005]

>>10589427
THANK YOU SO MUCH MAN
I love the prismrivers, bro

[Anonymous No.10590008]

Not even hard

[Anonymous No.10590024]

>>10590002
That doesn't prove convergence.
For example:
-1 <= sin(x) <= 1
Does sin(x) converge to anything as x goes to infinity?

[Anonymous No.10590025]

>>10590024
But the integrals of 1 and -1 don't converge

[Anonymous No.10590033]

>>10590025
Assume the -1 <= sin(x) <= 1 is what you get after integrating.
It seems like you are trying to use the squeeze theorem but the upper and lower bounds aren't converging to the same value.

[Anonymous No.10590039]

>>10590025
Do -sin(x^2) and sin(x^2) converge? It seems like they shouldn’t

[Anonymous No.10590040]

>>10590033
I'm not using the squeeze theorem. Just using the fact that if f <= g <= h, then their integrals are in the same order

[Anonymous No.10590043]

>>10590039
They do because they're just the Fresnel integrals as others have mentioned before

[Anonymous No.10590046]

>>10590040
My point is that just because a function is bounded doesn't mean it converges at infinity.
Oscillating functions like sin(x) are bounded but don't converge to a value as x goes to infinity.
Blowing up to infinity isn't the only way to diverge.

[Anonymous No.10590051]

>>10590043
Thanks, but I really don’t understand why sin(x^2) would converge if sin(x) doesn’t. Am I retarded?

[Anonymous No.10590058]

>>10590051
Ah, wait, I see it now

[Anonymous No.10590059]

Am I on the right track?

[Anonymous No.10590065]

>>10590051
Think of it as an alternating sum whose terms decrease to zero.
Part above x axis - Part below x axis + Part above x axis - ...

[Anonymous No.10590070]

>>10590002
putnam are such losers lmao how eaaaasy this shit it
prove the integrals are finite. (hint, they're not)
guys like you never make it

[Anonymous No.10590080]

>>10590059
No. Just write sin(x)sin(x^2) as
-(e^(ix) - e^(-ix))(e^(ix^2) - e^(-ix^2))/4
= -cos(x^2 + x)/2 + cos(x^2 - x)/2

Then use the argument from>>10590065
on each term.

[Anonymous No.10590087]

>>10590080
ah I see

[Anonymous No.10590116]

>>10590002
You forget that product sinx*sin(x^2) would change sign and may lose conditional convergence.

[Anonymous No.10590140]

>>10590046
>>10590070
>>10590116
Because sin(x^2) is integrable, [math] \forall \epsilon > 0, \exists B > 0: a, b > B \implies \int_{a}^{b} \sin(x^2) dx < \epsilon [/math].

Using [math] -\sin(x^2) \leq \sin(x) \sin(x^2) \leq \sin(x^2) [/math],
[math] -\epsilon < -\int_{a}^{b} \sin(x^2) dx \leq \int_{a}^{b} \sin(x) \sin(x^2) dx \leq \int_{a}^{b} \sin(x^2) dx < \epsilon [/math].
This shows that the integral of sin(x)sin(x^2) exists (because the sequence [math] a_n = \int_{0}^{n} \sin(x) \sin(x^2) dx [/math] is cauchy by the above argument)

[Anonymous No.10590179]

>>10590140
Use -sin(x^2) < sin(x^2)*sin(x^2) < sin(x^2)

[Anonymous No.10590192]

>>10589427
Knowing that sin(x^2) and cos(x^2) converge makes this problem ridiculously easy. sin(x)sin(x^2) = 0.5 (cos(x^2-x) - cos(x^2+x)) = 0.5(cos((x-0.5)^2-0.25) - cos((x+0.5)^2-0.25)).

[Anonymous No.10590211]

Oh okay I see where my proof is wrong:
-sin(x^2) < sin(x)sin(x^2) < sin(x^2) isn't actually true. If it were, then >>10590140 would hold. I made the elementary mistake of assuming that -1 < x < 1 implies -y < xy < y for all y (even though it only holds for positive y) Even the inequality in your example -sin(x^2) < sin(x^2)*sin(x^2) < sin(x^2) doesn't hold for the same reasons.

[Anonymous No.10590216]

>>10590211
Meant to reply to >>10590179

[Anonymous No.10590257]

>>10590080
But how do you show that it's a convergent alternating sum? I know by looking at the graph it's obvious...but is that good enough for a proof?

[Anonymous No.10590384]

>>10590257
You know the height of each "hump" is 1.
It's easy to show the width of each "hump" goes to zero as x goes to infinity.

[Anonymous No.10590593]

>>10590384
Is it enough to show that the terms go to zero, therefore it converges? What about the harmonic series? I think you need a few more steps

[Anonymous No.10590599]

Are the solutions to these problems posted anywhere?

[Anonymous No.10590608]

>>10590593
You just need to show they are decreasing in absolute value.
https://en.wikipedia.org/wiki/Alternating_series_test

[Anonymous No.10590628]

>>10590608
Ah, I see. Calc II is coming back to me

[Anonymous No.10590633]

>>10590140
proof it's integrable?
you're literally just writing what it means it's integrable, not proving it

[Anonymous No.10590855]

>>10590384
> It's easy to show the width of each "hump" goes to zero as x goes to infinity.
If it's so easy than, show me. Inb4 look at the graph.

[Anonymous No.10590887]

>>10590855
n/m it was easy.

[math] \sin(x^2) = 0 [/math]
[math] x_{n}^{2} = n \pi [/math]
[math] x_n = \sqrt{n \pi} [/math]
[math] \Delta x_n = x_{n+1} - x_n [/math]
[math] \lim\limits_{n \to \infty} \Delta x_n = 0 [/math]

[http://math.harvard.edu/putnam(...) !!84u21X0KjW+ No.10591662]

Is there anything you'd like me to change about these daily putnam threads? I'm wide open to suggestions.

[Anonymous No.10592096]

>>10591662
Have Sex

[Anonymous No.10592106]

>>10591662
Post selfies

[Anonymous No.10592144]

>>10591662
How about making two threads?
One with preliminary questions/hints to help those who need it
And the usual one for those who like to be challenged
Posters also need to use spoilers

[Anonymous No.10592296]

>>10589427
Since Fresnel Integrals make this problem a joke, here’s another solution:
Let out integral be from a to b, finding limits a->0 and b->infinity will give desired result.
Integral(a,b) sinx*sin(x^2) dx= -(sinx*cos(x^2))/2x b|a + integral(a,b) ((xcosx - sinx)cos(x^2))/2x^2 dx.
(sinx*cos(x^2))/2x has both finite limits for 0 and inf so we have to proof that integral ((xcosx - sinx)cos(x^2))/2x^2 dx converges.
Do it the same way:
Integral(a,b) ((xcosx - sinx)cos(x^2))/2x^2 dx = ((xcosx - sinx)sin(x^2))/4x^3 b|a + integral(a,b) (4x^4*sinx*sin(x^2) - 12x^2*(xcosx - sinx)*sin(x^2))/16x^6.
Again
((xcosx - sinx)sin(x^2))/4x^3 has finite limits for zero and inf. Look at the integral, it clearly converges in b goes to infinity but it may diverge when a goes to zero. Lets check this limit:
1. (4x^4*sinx*sin(x^2))/16x^6 = sinx*sin(x^2)/4x^2. It’s sinx/x limit and Lim x->0 sinx*sin(x^2)/4x^2 = 0
2. 12x^2*(xcosx - sinx)*sin(x^2))/16x^6 = 3(xcosx - sinx)*sin(x^2))/4x^4. Using LHopital’s rule we can see that this limit is also zero.
Hence we can see that the integral converges.

[Anonymous No.10592304]

>>10589427
who is putnam?