12.10. EXERCISES 297
x
x = γy
It turns out that in this example, and many other examples, the circular part of thecontour integral, denoted here with
∫CR, will converge to 0 as R→ ∞ and so
limR→∞
12π
∫ R
−Re(γ+iy)tL g(γ + iy)dy =
12π
2πi(−i)(sum of residues) = sum of residues
In this example,
limz→i
ezt
z+ i=
12i
eit , limz→−i
ezt
(z− i)= e−it 1
−2i
Thus, these add to12i
eti +
(e−it 1−2i
)=
12
ie−it − 12
ieit = sin t
Thus the Bromwich improper integral is 12π
2π sin t = sin t. We just found an inverse Laplacetransform. In general, this illustrates the following procedure which works under fairlygeneral conditions.
Procedure 12.9.3 Let F (s) satisfy |F (z)|<C/ |z|α for some α > 0 for all |z| largeenough with F (z) analytic except for finitely many poles. To find f (t) such that L f = F,
f (t) =(
sum of residues at the poles of F (z)etz)12.10 Exercises1. Generalize the Riemann Lebesgue lemma to show that if f is piecewise continuous
on finite intervals and in L1 (R) , then limn→∞
∫∞
−∞f (t)sin(nt + k)dt = 0. Here k is
any constant.
2. Show that if f is in L1 (R) , then limn→∞
∫∞
−∞f (t)cos(nt + k)dt = 0.
3. In fact, all you need in the above problems is to assume that f is Riemann integrableon finite intervals and that limR→∞
∫ R−R f (t)dt exists. Show this. Hint: This mainly
requires fussing over whether you end up with something Riemann integrable giventhat f is when you do certain things. I have presented it for piecewise continuousfunctions because that is where the interesting application resides and it is obviousthat everything stays piecewise continuous. If you want to do these things right, youneed to be using the Lebesgue integral anyway.
4. limR→∞
∫ R−R
t1+t2 dt = 0 but limn→∞
∫ R2
−Rt
1+t2 dt = ∞. Show this.
5. Here are some functions which are either even or odd. Use 12.15, 12.16 to representthese functions and thereby obtain amazing integration formulas. Be sure to dealwith the situation at jumps.