(u/v)''=(u''v-v''u)/(v^2)
Ummm... seems to doesn''t work
Is this integrable?
That was my first thought...but try this formula on your result
(u/v)''=(u''v-v''u)/(v^2)
Ummm... seems to doesn''t work
(u/v)''=(u''v-v''u)/(v^2)
Ummm... seems to doesn''t work
the u/v thing is when differentiating.
the differential of cos nx is -nsin nx, so the integral of
sin nx is (-cos nx)/n
(Where n is a constant)
I don''t have time to go into the laws of integration
the differential of cos nx is -nsin nx, so the integral of
sin nx is (-cos nx)/n
(Where n is a constant)
I don''t have time to go into the laws of integration
i have tried to integrate that function in mathcad and the result
had some fourier transorm shit in it and pi raised to some weird power, this is a very complex intergration
had some fourier transorm shit in it and pi raised to some weird power, this is a very complex intergration
mm... i thought this was simple. I''ll go over it again lter on paper - I cannot think properly without a ''scratchpad''
my notes in class look awful
my notes in class look awful
ok, first off
the integral of sin(x^2) IS NOT (-cos(x^2))/(2x), you may be able to differentiate back to sin(x^2) but (-cos(x^2))/(2x) does not exist at 0 but the integral of sin(x^2) at 0 does. (check: expand sin(x^2) as a maclurin series and then integrate .. 0 does exist). Also if i remember correctly sin(x^2) does not have integral within the basic functions of calculus ... so you must expand it as a Taylor series then integrate ...
Second,
I would go with declspec suggestion and expand sin(x^2) as maclurin series then intergrate. Now it's up to you on how many terms you need, more terms = more accurate approximation of intergral of sin(x^2).
hope that helped.
jp.
==============================================
I feel like a kid in some kind of store...
==============================================
www.thejpsystem.com
[edited by - loserkid on May 28, 2002 4:11:03 PM]
the integral of sin(x^2) IS NOT (-cos(x^2))/(2x), you may be able to differentiate back to sin(x^2) but (-cos(x^2))/(2x) does not exist at 0 but the integral of sin(x^2) at 0 does. (check: expand sin(x^2) as a maclurin series and then integrate .. 0 does exist). Also if i remember correctly sin(x^2) does not have integral within the basic functions of calculus ... so you must expand it as a Taylor series then integrate ...
Second,
I would go with declspec suggestion and expand sin(x^2) as maclurin series then intergrate. Now it's up to you on how many terms you need, more terms = more accurate approximation of intergral of sin(x^2).
hope that helped.
jp.
==============================================
I feel like a kid in some kind of store...
==============================================
www.thejpsystem.com
[edited by - loserkid on May 28, 2002 4:11:03 PM]
Mine was analytic. Not that mine was the only analytic one.
[edited by - LilBudyWizer on May 24, 2002 2:34:42 PM]
[edited by - LilBudyWizer on May 24, 2002 2:34:42 PM]
Author
Thanks guys, this has been informative. As it turns out my teacher is going to let me do the integral with a maclauren series expansion.. so that''ll make it nice and clean. And as it turns out for the required domain I only need like 4 terms of the series to be accurate enough.
Thanks!
Thanks!
Basically, if you find the integral of f(x), but when differentiating back to f(x) you cancel out an x term... that usually means you don''t have the proper integral in the first place. Cancelling out x terms is sometimes bad, especially since x/x is indeterminate when x = 0. Watch out!
As many people have said the integral 0∫a sin(x2) dx is an elementary function called the Fresnel sin function. These are evaluated numerically and look quite pretty when plotted.
For the specific case 0∫infinity sin(x2) dx you can actually evaluate it.
Make the change of variables u = x^2 and then use the formula sin(u) = 0.5*i(e-iu - eiu)
You then make two more changes of variables on the two integrands containing e-iu and eiu and you end up with this
0.5 * cos(PI/4) * 0∫infinityv-0.5e-v dv
The integral on the right is Gamma(0.5) = sqrt(PI) where Gamma is the generalised factorial function.
So the whole thing is 0.5 * sqrt(PI/2).
Don't know if this helps but it's kinda neat
[edited by - sQuid on May 28, 2002 11:33:47 PM]
For the specific case 0∫infinity sin(x2) dx you can actually evaluate it.
Make the change of variables u = x^2 and then use the formula sin(u) = 0.5*i(e-iu - eiu)
You then make two more changes of variables on the two integrands containing e-iu and eiu and you end up with this
0.5 * cos(PI/4) * 0∫infinityv-0.5e-v dv
The integral on the right is Gamma(0.5) = sqrt(PI) where Gamma is the generalised factorial function.
So the whole thing is 0.5 * sqrt(PI/2).
Don't know if this helps but it's kinda neat
[edited by - sQuid on May 28, 2002 11:33:47 PM]
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