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<math>\int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw =1 \ \ \text{if} \ \ 0 <v < 2 \ \ \ \text{and } 0 \ \ \ \text{if} \ v >2 </math>. | <math>\int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw =1 \ \ \text{if} \ \ 0 <v < 2 \ \ \ \text{and } 0 \ \ \ \text{if} \ v >2 </math>. | ||
| − | + | Use the second and the last equations. Then, you will get the answer. | |
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Revision as of 11:22, 4 November 2013
Homework 10 collaboration area
From Jake Eppehimer:
I must be doing something wrong on problem 2 of Lesson 31. There doesn't seem to be a way to solve the integral if you use formula (1b) with the answer from problem 1. Any tips?
Also, I don't understand how to do number 5 on Lesson 29. Any help would be appreciated.
Response from Mickey Rhoades Mrhoade
I get 2/pi times a bunch of sinc function integrals which have to be evaluated with the Dirichlet Integral. I can use Table 11.10 relationship #10 and get the answer without chugging out all the integrals. ---
From Jake:
What page is this table on? I'm not seeing anything. Thanks.
From Andrew:
Table 1 in 11.10 on Page 534 is a table of Fourier Cosine Transforms.
From Eun Young:
From the lesson 30 lecture note, we know that $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if -1<v<1 and 0 otherwise. Consider only positive v. Then, $ \int_0^{\infty} \frac{2 \sin w }{\pi w} \cos wv \ \ dw = 1 $ if 0 <v<1 and 0 if v>1.
From this, we can compute $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw . $
Let 2w = t. Then, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw = \int_0^{\infty} \frac{ 2 \sin t }{ \pi \frac t 2} cos (t \frac v 2 ) \frac{dt}{2} = \int_0^{\infty} \frac{2 \sin t}{ \pi t} cos(\frac v 2 t) \ \ dt = 1 \ \ \text{if} \ \ 0 < \frac v 2 < 1 \ \ \ \text{and } 0 \ \ \ \text{if} \ \ \frac v 2 >1 $.
Thus, $ \int_0^{\infty} \frac{2 \sin 2w }{\pi w} \cos wv \ \ dw =1 \ \ \text{if} \ \ 0 <v < 2 \ \ \ \text{and } 0 \ \ \ \text{if} \ v >2 $.
Use the second and the last equations. Then, you will get the answer.
