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How the heck do I integrate sin(1/x) ?

by Matt Jacques
Tags: heck, integrate, sin1 or x
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Matt Jacques
#1
Feb18-04, 06:06 PM
P: 80
I tried parts by integration but Im caught in an endless loop of ever growing in complexity integrals! I must be missing something.
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matt grime
#2
Feb18-04, 06:21 PM
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is that an indefinite or definite integral?
HallsofIvy
#3
Feb18-04, 08:24 PM
Math
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Thanks
PF Gold
P: 39,533
Do you have any reason to believe its anti-derivative is an elementary function?

PrudensOptimus
#4
Feb18-04, 11:46 PM
P: 640
How the heck do I integrate sin(1/x) ?

easy man

here's the answer:


-cosintegral[1/x] + xSin[1/x]
Matt Jacques
#5
Feb19-04, 06:47 AM
P: 80
I got that, too. No way to further simplify?
matt grime
#6
Feb19-04, 07:21 AM
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apart from that the integral of 1/x is log(x) you mean?
NateTG
#7
Feb19-04, 10:02 AM
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If you're desperate, you could try working out a Taylor/Mclaurin series for it, and seeing if the integral of that is recognizable.
Spectre5
#8
Feb20-04, 04:12 PM
P: 186
You can use a Maclaurin series to evaluate (or at least approximate) it...knowing that

[tex]sin(x)=\sum_{n=0}^{\infty}\frac{(-1)^nx^{(2n+1)}}{(2n+1)!}[/tex]


you can replace x with 1/x and integrate to get:


[tex]\int sin(\frac{1}{x})=\sum_{n=0}^{\infty}\frac{(-1)^{n-1}}{2(2n+1)!x^{2n}}[/tex]
PrudensOptimus
#9
Feb20-04, 07:10 PM
P: 640
Originally posted by matt grime
apart from that the integral of 1/x is log(x) you mean?

Wrong. ∫1/x dx = ln |x| + C.

∫1/(x(ln 10)) dx = log |x| + C.
master_coda
#10
Feb20-04, 07:13 PM
P: 678
Originally posted by PrudensOptimus
Wrong. ∫1/x dx = ln |x| + C.

∫1/(x(ln 10)) dx = log |x| + C.
When a mathematician says "log" they are generally talking about the natural logarithm.
NateTG
#11
Feb20-04, 08:32 PM
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Originally posted by master_coda
When a mathematician says "log" they are generally talking about the natural logarithm.
Right, and the rest of the time they usually mean [tex]log_2[/tex]
but anything other than [tex]log_e[/tex] gets a base.
master_coda
#12
Feb20-04, 11:56 PM
P: 678
Originally posted by NateTG
Right, and the rest of the time they usually mean [tex]log_2[/tex]
but anything other than [tex]log_e[/tex] gets a base.
I don't see too many mathematicians refer to [itex]\log_2[/itex] as [itex]\log[/itex].
NateTG
#13
Feb21-04, 12:22 AM
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It's typically for math/cs tpe situations and usually only applies to situations where hte base is not particuarly important.
matt grime
#14
Feb21-04, 03:12 AM
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Originally posted by PrudensOptimus
Wrong. ∫1/x dx = ln |x| + C.

∫1/(x(ln 10)) dx = log |x| + C.

yes, i did omit the modulus sign, however you should probably be told that log always means base e. This is completely standard in mathematics, and just one more thing they misteach at high school


After all what other base would you possibly want?
curiousbystander
#15
Feb24-04, 11:12 AM
P: 20
This might help too:

sin(1/x) is an odd function (meaning f(-x) = -f(x)).

The definite integral of any odd function on the interval [-a,a] is 0.
matt grime
#16
Feb24-04, 12:04 PM
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one generally wouldn't integrate over a region where the function is not defined. (no choice at zero can make it continuous, interestingly enough, not that that's either here or there, and not that any choice would make the integral be anything but zero anyway, though 0 is the only choice that keeps it a genuine odd function.)
curiousbystander
#17
Feb24-04, 05:17 PM
P: 20
I should have been more careful when answering, but isn't the integral still well defined since {0} is a set of measure 0?
NateTG
#18
Feb24-04, 05:28 PM
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Do you mean to use Lebesgue integration?

[tex]\lim_{x \rightarrow 0}[/tex] might also not exist and thus cause problems.


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