
#19
Feb1312, 07:23 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525

Hint:
$$\int_0^k \cos kp\,dk = \frac{\sin kp}{p}$$ 



#20
Feb1312, 08:03 PM

PF Gold
P: 3,173





#21
Feb1512, 01:13 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525

I edited my hint to make it more suggestive. Use it to evaluate the integral in post #17.




#22
Feb1512, 01:17 PM

PF Gold
P: 3,173

Ok vela, I've been helped on that particular integral (though I still have doubts) by JJacquelin.
I think the main idea was to consider the integrand as a function of k, then differentiate it with respect to k. Then integrate this result (from 0 to infinity) with respect to p (it's slightly less "ugly" than the original integral, though I'm still unsure how to perform that integral but I will ask for help). Then integrate this result with respect to k and take the constant of integration equal to 0. All in all this gives [itex]\int _0 ^{\infty }\frac{\sin (kp)e^{px}}{p}dp=\arctan \left ( \frac{k}{x} \right )[/itex]. Hence [itex]u(x,y)=\frac{1}{\pi} \arctan \left ( \frac{y+a}{x} \right )+\frac{1}{\pi} \arctan \left ( \frac{ay}{x} \right )[/itex]. I'd like to know whether your method is the same. If it's a different, I'd like to try it but I don't see how I can use your hint so far. Edit: I just see that you just replied to my last post. I'll check this out. 



#23
Feb1512, 01:19 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525

It's slightly different but yields the same result.




#24
Feb1512, 01:32 PM

PF Gold
P: 3,173

[itex]I=\int _0 ^{\infty } \int _0^{k} \cos (kp)dke^{xp}dp=\int _0^k \int _0 ^{\infty } \cos (kp)e^{px}dpdk[/itex]. I'm stuck on that integral with respect to p (integral by parts failed and I don't see any useful substitution so I used Wolfram Alpha for it). [itex]I=\int _0 ^k \frac{e^{xp}[k \sin (kp)x \cos (kp)]}{k^2+x^2} \big  _0 ^{\infty } dk =\int _0 ^k \frac{x}{k^2+x^2}dk=\arctan \left ( \frac{k}{x} \right )[/itex].
Edit: Apparently there are at least 2 ways to tackle down that integral (the one I used wolfram alpha). A double integration by parts or a rewriting of the cos term in function of exponentials. I'll be looking forward this. Thank you very much for all vela. So the answer is post #22 is correct, right? 



#25
Feb1512, 01:52 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525





#26
Feb2112, 11:14 PM

PF Gold
P: 3,173

Instead of starting a new thread about sine/cosine Fourier transform I'll ask here.
Basically I wonder why the problem statement made it clear to take a cosine Fourier transform instead of just a Fourier transform. Is it because it is obvious that the solution to the PDE, u(x,y) is an even function for any fixed x? So that a common Fourier transform would lead to a cosine Fourier transform anyway. P.S.: Just for fun, I think I've plotted the solution function (see post 9 and maybe 10 of http://www.physicsforums.com/showthread.php?t=577993). 



#27
Feb2212, 05:41 AM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525

It may have been simply for pedagogical reasons.




#28
Feb2212, 02:19 PM

PF Gold
P: 3,173





#29
Feb2212, 03:16 PM

Emeritus
Sci Advisor
HW Helper
Thanks
PF Gold
P: 11,525

Yes. That's where the Fourier cosine transform comes from, right? You plug the function into the regular Fourier transform integral and then simplify it using the fact the function is even.



Register to reply 
Related Discussions  
Fourier transform and the heat equation, don't understand the provided answer  Calculus & Beyond Homework  0  
Fourier Transform to solve heat equation in infinite domain  Calculus & Beyond Homework  7  
Cosine Fourier Transform Spectroscopy  Calculus  0  
Solving Nonhomogeneous Heat Equation with Fourier Transform  Calculus & Beyond Homework  2  
Fourier Transform of cosine and rect  Calculus & Beyond Homework  1 