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Interesting problem.

lima0asinxxdx \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

It has a cool answer.
Reply 1
Original post by Louisb19
lima0asinxxdx \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

It has a cool answer.


Off the top of my head, I can think of two ways to do this:

DUTIS:

Spoiler


Laplace transforms:

Spoiler

Reply 2
Original post by Zacken
Off the top of my head, I can think of two ways to do this:

DUTIS:

Spoiler

Laplace transforms:

Spoiler



It's a pretty cool answer isn't it.
Reply 3
Original post by Louisb19
It's a pretty cool answer isn't it.


Indeed, can you hazard a guess as to what sinxxdx\displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x} \, \mathrm{d}x is?
Reply 4
Original post by Zacken
Indeed, can you hazard a guess as to what sinxxdx\displaystyle \int_{-\infty}^{\infty} \frac{\sin x}{x} \, \mathrm{d}x is?


I'd guess since f(x)=sinxx=f(x)=sin(x)x f(x) = \dfrac{\sin x}{x} = f(-x) = \dfrac{\sin (-x)}{-x} (I think?)

It would be 20sinxxdx=π 2\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} dx = \pi
Reply 5
Original post by Louisb19
I'd guess since f(x)=sinxx=f(x)=sin(x)x f(x) = \dfrac{\sin x}{x} = f(-x) = \dfrac{\sin (-x)}{-x} (I think?)

It would be 20sinxxdx=π 2\displaystyle \int_{0}^{\infty} \frac{\sin x}{x} dx = \pi


Precisely. sin(x)=sinx\sin (-x) = -\sin x so f(x)f(x) is an even function.
So how about

0sin2xx2dx\displaystyle \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, \mathrm{d}x ?
Reply 7
Original post by Zacken
Off the top of my head, I can think of two ways to do this:

DUTIS:

Spoiler


Laplace transforms:

Spoiler



the Laplace approach is not quite correct
Reply 8
Original post by TeeEm
the Laplace approach is not quite correct


Is this correct (sorry, my Laplace is horridly rusty):

L{sinxx}=0L{sint}ds=01s2+1ds=π2\displaystyle L\left\{\frac{\sin x}{x}\right\} = \int_0^{\infty} L\{\sin t\}\, \mathrm{d}s = \int_0^{\infty} \frac{1}{s^2 + 1} \, \mathrm{d}s = \frac{\pi}{2}
(edited 8 years ago)
Reply 9
Original post by Gregorius
So how about

0sin2xx2dx\displaystyle \int_{0}^{\infty} \frac{\sin^2 x}{x^2} \, \mathrm{d}x ?


Is this just using

Spoiler

Reply 10
Original post by Louisb19
lima0asinxxdx \lim_{a\to\infty} \displaystyle\int_{0}^{a} \dfrac{sinx}{x}dx

It has a cool answer.


You might like this. You come up with quite a few interesting problems, so you can post them there in the future! :smile:
Reply 11
Original post by Zacken
Is this correct (sorry, my Laplace is horridly rusty):

L{sinxx}=0L{sint}ds=01s2+1ds=π2\displaystyle L\left\{\frac{\sin x}{x}\right\} = \int_0^{\infty} L\{\sin t\}\, \mathrm{d}s = \int_0^{\infty} \frac{1}{s^2 + 1} \, \mathrm{d}s = \frac{\pi}{2}


I just got back ...
food first!
Then I will write and post the method
Reply 12
Original post by TeeEm
I just got back ...
food first!
Then I will write and post the method


Bon appetit! :smile:
Reply 13
Original post by Zacken
Bon appetit! :smile:


here is the one and another 3 for the interest of whoever cares

1.jpg
Attachment not found
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edexcel sending.jpg


I posted earlier another integral problem here
http://www.thestudentroom.co.uk/showthread.php?t=3327325

Maybe this is easy to be done now
Reply 14
Original post by TeeEm
here is the one and another 3 for the interest of whoever cares

1.jpg
Attachment not found
Attachment not found
edexcel sending.jpg


I posted earlier another integral problem here
http://www.thestudentroom.co.uk/showthread.php?t=3327325

Maybe this is easy to be done now


Many thanks!
Original post by Zacken
Is this just using

Spoiler



Looks about right; what is amusing in the context of the original question is the actual value of the answer...

Unfortunately the pattern does not continue for

0sinnxxndx \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}x

for n in general...
(edited 8 years ago)
Reply 16
Original post by Gregorius
Looks about right; what is amusing in the context of the original question is the actual value of the answer...Unfortunately the pattern does not continue for0sinnxxndx \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}xfor n in general...

I was just about to ask about that. :lol:

Interestingly enough we do have that becomes 3π8\frac{3\pi}{8}
(edited 8 years ago)
Reply 17
Original post by Gregorius
Looks about right; what is amusing in the context of the original question is the actual value of the answer...

Unfortunately the pattern does not continue for

0sinnxxndx \displaystyle \int_{0}^{\infty} \frac{\sin^n x}{x^n} \, \mathrm{d}x

for n in general...


I'll be back in half an hour or so, I'm going to play around with that integral and see if I can any nice recurrences or patterns! Thanks for this!

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