Kenneth Kuttler

298 CHAPTER 12. SERIES AND TRANSFORMS

(a) f (x) = X[−1,1] (x). Thus f (x) = 1 on [−1,1] and is 0 elsewhere. Show that

limR→∞

2π

∫ R

0cos(tx)

(1t

sin t)

dt =

 1 if x ∈ (−1,1)1/2 if |x|= 10 if |x|> 1

Isn’t this an amazing result?

(b) f (x) = cos(x) on(−π

2 ,π

)and 0 if |x|> π

2 .

(d) f (x) = cos(x) on (−π,π) and 0 if |x|> π.

(e) f (x) = e−x if x > 0,−e−|x| if x < 0.

6. Find limR→∞

∫ 1−1

sin(πt)sin(Rt)t2 dt. Hint: Consider the following steps. Using the ver-

sion of Fubini’s theorem from Theorem 9.9.4 on Page 214∫ 1

−1

sin(πt)sin(Rt)t2 = 2

∫ 1

sin(πt)sin(Rt)t2 dt

= 2∫ 1

sin(πt)t

∫ R

0cos(yt)dydt = 2

∫ R

∫ 1

sin(πt)t

cos(yt)dtdy

Now the Fourier cosine transform for the even function x→ sin(πx)x X[−1,1] ≡ f (x) is

limR→∞

2π

∫ R

0cos(xy)

∫ 1

sin(πt)t

cos(yt)dtdy = f (x)

This equals the value of this function at x. It looks like what this problem is askingis π f (0) . What is f (0)?

7. Do something similar to the above problem in order to compute

limR→∞

∫ 1

sin(Rx)√x

Hint:∫ 1

0sin(Ry)√

y dy =∫ 1

0√

y∫ R

0 cos(ty)dtdy =∫ R

0∫ 1

0√

ycos(ty)dydt. Isn’t this a case

of the Fourier cosine transform for the even function x→√|x|?

8. In example 12.9.2, verify the details. In particular, show that the contour integralover the circular part of the contour converges to 0 as R→ ∞. Explain why γ couldbe any positive number in this case. In general, you just need γ to be large enoughso that the contour encloses all the poles and the function for which you are trying tofind the inverse Laplace transform, needs to be dominated by C/ |z|α for some α > 0for all |z| large enough. Try to show this. It is a little technical but only involveselementary considerations. Thus verify Procedure 12.9.3 is valid by showing thatthe part of the contour integral over the circular part converges to 0 if the assumedestimate holds.

9. Find inverse Laplace transforms using the method of residues in Procedure 12.9.3for b

(s−a)2+b2 , s−a(s−a)2+b2 , and the other entries of the table in Problem 8 on Page 246.

298CHAPTER 12. SERIES AND TRANSFORMS(a) f(x) = 2-11 (x). Thus f (x) = 1 on [—1, 1] and is 0 elsewhere. Show that2 PR , 1 if x € (-1,1)lim — / cos (tx) (sin) dt = 1/2 if |x| =1Roe J0 f Oif |x| >1Isn’t this an amazing result?(b) f(x) = €0s (x) on (—§, ¥) and Oif |x| > 4(c) f(x) = me ) on (—22, 27) and 0 if |x| > “om.(d) f (x) =cos(x) on (—2, 7) and 0 if |x| > z.(x)(e) f(x) =e ifx > 0,-e “Hl ifx <0.6. Find limr_5.o fi in Hint: Consider the following steps. Using the ver-sion of Fubini’s theorem from Theorem 9.9.4 on Page 214! sin (ar) sin (Rr) 9 ! sin (zr) sin (Rt) dtat2 t2lg R R pl gj= 2 | ina) | cos (yt) dydt = 2/ | sn os (yt) dtdy0 0 0 JotNow the Fourier cosine transform for the even function x > sin(as) Asy=fh (x) is2 R lg tlim = | cos (xy) | sin (1) cos (yt) dtdy = f (x)R>0 7 Jo 0 tThis equals the value of this function at x. It looks like what this problem is askingis 1 f (0). What is f (0)?Do something similar to the above problem in order to computelelim [ sin (Rx) dxRoo JO VxHint: fo sin mn dy =fy Vy Io cos (ty) dtdy = Se fo ,/ycos (ty) dydt. Isn’t this a caseof the Fourier cosine transform for the even function x + \/|x|?. In example 12.9.2, verify the details. In particular, show that the contour integralover the circular part of the contour converges to 0 as R + co. Explain why y couldbe any positive number in this case. In general, you just need y to be large enoughso that the contour encloses all the poles and the function for which you are trying tofind the inverse Laplace transform, needs to be dominated by C/ |z|* for some a > 0for all |z| large enough. Try to show this. It is a little technical but only involveselementary considerations. Thus verify Procedure 12.9.3 is valid by showing thatthe part of the contour integral over the circular part converges to 0 if the assumedestimate holds.Find inverse Laplace transforms using the method of residues in Procedure 12.9.3for wae ; te , and the other entries of the table in Problem 8 on Page 246.