Can the Limits of Integration be Changed in the Fresnel Integral?

[Nanie]

I need help for this integral


cosx cosx^{2}

 

[gabbagabbahey]

I need help for this integral cosx cosx^{2}

Do you mean
\int cos(x) cos^2(x) dx
or
\int cos(x) cos(x^2) dx
?

 

[HallsofIvy]

Whatever you mean, what have you tried?

 

[Nanie]

cos(x)cos(x^2)

 

[Defennder]

I don't think the anti-derivative can be expressed in elementary functions.

 

[gabbagabbahey]

cos(x)cos(x^2)

You will need to integrate this by-parts and express your answer in terms of Fresnel Integrals which are defined here: http://http://en.wikipedia.org/wiki/Clothoid#Cornu_spiral"

 

[Nanie]

thanks... sorry for my english...I hope that can understand me...

This integral <br /> \int_{1} ^{\infty} cos(x)cos(x²) dx

I have to say that is convergent or divergent. The proffessor gave us a hint (Fresnel Integral)...but I don't know how to use it, He don't discuss it in class . I understand everything but the cos(x²) jeje I don't know how to integrate.

Thanks gabbagabbahey for the link...

 

[gabbagabbahey]

thanks... sorry for my english...I hope that can understand me...

This integral <br /> \int_{1} ^{\infty} cos(x)cos(x²) dx

I have to say that is convergent or divergent. The proffessor gave us a hint (Fresnel Integral)...but I don't know how to use it, He don't discuss it in class . I understand everything but the cos(x²) jeje I don't know how to integrate.

Thanks gabbagabbahey for the link...

Okay, this is going to be ugly, but here it goes:

cos(x)cos(x^2)=\frac{cos(x^2-x)+cos(x^2+x)}{2}= \frac{cos \left( x^2-x+\frac{1}{4} - \frac{1}{4} \right)+cos \left( x^2+x+\frac{1}{4} - \frac{1}{4} \right) }{2} = \frac{cos \left( \left( x-\frac{1}{2} \right) ^2 - \frac{1}{4} \right)+cos \left( \left( x+\frac{1}{2} \right) ^2 - \frac{1}{4} \right) }{2}


= \frac{ cos \left( \left( x - \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x - \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) +cos \left( \left( x + \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x + \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) }{2}


\Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx

= \frac{1}{2} \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x - \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) +cos \left( \left( x + \frac{1}{2} \right) ^2 \right) cos \left( \frac{1}{4} \right) +sin \left( \left( x + \frac{1}{2} \right) ^2 \right) sin \left( \frac{1}{4} \right) dx

= \frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x - \frac{1}{2} \right) ^2 \right)dx + cos \left( \frac{1}{4} \right) \int_1^{\infty} cos \left( \left( x + \frac{1}{2} \right) ^2 \right) dx

\left{+ sin \left( \frac{1}{4} \right) \int_1^{\infty} sin \left( \left( x + \frac{1}{2} \right) ^2 \right) dx \right]


...

 

[gabbagabbahey]

...

Make the following substitutions:

u \equiv x-\frac{1}{2} , \quad v \equiv x+ \frac{1}{2} \quad \Rightarrow du=dv=dx , \quad u: \frac{3}{2} \rightarrow \infty , \quad v: \frac{1}{2} \rightarrow \infty

\Rightarrow \int_1^{\infty} cos(x)cos(x^2)dx
= \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} cos(u^2)du + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{3}{2}}^{\infty} sin(u^2)du + \frac{1}{2} cos \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} cos(v^2)dv + \frac{1}{2} sin \left( \frac{1}{4} \right) \int_{\frac{1}{2}}^{\infty} sin(v^2)dv

...

 

[gabbagabbahey]

...
Now using the definition of the Fresnel sine and cosine integrals, S(t),\quad C(t) :

S(t) \equiv \int_0^t sin(q^2)dq , \quad C(t) \equiv \int_0^t cos(q^2)dq

we get:

\int_1^{\infty} cos(x)cos(x^2)dx = \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{3}{2} \right) \right) + \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{3}{2} \right) \right) + \frac{1}{2} cos \left( \frac{1}{4} \right) \left( C(\infty)-C \left( \frac{1}{2} \right) \right)
+ \frac{1}{2} sin \left( \frac{1}{4} \right) \left( S(\infty)-S \left( \frac{1}{2} \right) \right)
...

 

[gabbagabbahey]

...And using the fact that
S(\infty) = C(\infty) =\frac{\sqrt{\pi}}{8}
this becomes:

\int_1^{\infty} cos(x)cos(x^2)dx = -\frac{1}{2} \left[ cos \left( \frac{1}{4} \right) \left( C \left( \frac{3}{2} \right) + C \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) + sin \left( \frac{1}{4} \right) \left( S \left( \frac{3}{2} \right) +S \left( \frac{1}{2} \right) -\frac{\sqrt{\pi}}{4} \right) \right]<br />

 

[Nanie]

iah...very ugly...wow!

thanks thanks... jeje now I will try to understand it!

 

[medinap]

im nanies roommate and we take the class together and the problem is that i don't understand what u did in the first part of this problem and don't know where the cosX^2 - X come from. also i don't understand why u divided everything by 2...please help me

 

[gabbagabbahey]

im nanies roommate and we take the class together and the problem is that i don't understand what u did in the first part of this problem and don't know where the cosX^2 - X come from. also i don't understand why u divided everything by 2...please help me

I used the following Trig Identity:

cos(A)cos(B)=\frac{cos(A-B)+cos(A+B)}{2}

 

[medinap]

ok thank you soo much! the thing is that i have being trying it since tuesday! but thanks

 

[Nanie]

THANK YOU!

 

[Nanie]

a doubt come into my mind when i was doing the calculus exercise. ...i need to know if in the fresnel integral i can used any limit of integration or only from 0 to infinite...

 

[gabbagabbahey]

a doubt come into my mind when i was doing the calculus exercise. ...i need to know if in the fresnel integral i can used any limit of integration or only from 0 to infinite...

If the limits are zero to infinty, then you have S(infinty), but if the limits are 0 to a, you have S(a) as long is a is positive, the fresnel integrals converge.