11.7. EXERCISES 275

24. Show using methods from real analysis that for b≥ 0,∫

0 e−x2cos(2bx)dx =

√π

2 e−b2

Hint: Let F (b)≡∫

0 e−x2cos(2bx)dx−

√π

2 e−b2. Show F (0) = 0. Recall that∫

0e−x2

dx =12√

π.

Explain using a modification of Problem 48 on Page 224 why

F ′ (b) =∫

0−2xe−x2

sin(2bx)dx+2b√

π

2e−b2

F ′ (b) = 2b(∫

0e−x2

cos(2bx)dx+√

π

2e−b2

)

= 2b(

F (b)+√

π

2e−b2

+

√π

2e−b2

)= 2bF (b)+

√π2be−b2

Now use the integrating factor method for solving linear differential equations frombeginning differential equations to solve the ordinary differential equation. If youhave not seen this method, it is just this. To solve y′+ f (x)y = g(x) , multiply bothsides by eF(x) where F ′ (x) = f (x) . This reduces the left side to d

dx

(eF(x)y

). Thus

ddb

(e−b2

F (b))=√

π2be−2b2Then do

∫db to both sides and use that F (0) = 1

2√

π .

25. You can do the same problem as above using contour integration. For b > 0, use thecontour which goes from −a to a to a+ ib to −a+ ib to −a. Then let a→ ∞ andshow that the integral of e−z2

over the vertical parts of this contour converge to 0.Hint: For z = x+ ib,e−z2

= e−(x2−b2+2ixb) = eb2e−x2

(cos(2xb)+ isin(2xb)) .

26. Consider the circle of radius 1 oriented counter clockwise.∫

γz−6 cos(z)dz =?

27. The circle of radius 1 is oriented counter clockwise. Evaluate∫

γz−7 cos(z)dz.

28. Find∫

02+x2

1+x4 dx.

29. Find∫

0x1/3

1+x2 dx.

30. Suppose f is analytic on an open set U and α ∈U. Define

g(z)≡{ f (z)− f (α)

z−αif z ̸= α

f ′ (α) if z = α

show that g is analytic on U .

31. Let γ∗ be a C1 oriented closed curve and let α /∈ γ∗ Then

n(γ,α)≡ 12πi

∫γ∗

1z−α

dz = m ∈ N

This is called the winding number. Hint: Now let g(t) ≡∫ t

a1

γ ′(s)γ(s)−α

dt. Show thatddt

(e−g(t) (γ (t)−α)

)= 0. Explain why this requires e−g(t) (γ (t)−α) is a constant.

Since γ parametrizes a closed curve, argue that g(bk) = 2mπi for an integer m.

11.7. EXERCISES 27524. Show using methods from real analysis that for b > 0, fy e 2” cos (2bx) dx = vie25.26.27.28.29.30.31.Hint: Let F (b) = Jy e e-* cos (2bx) dx — VE Show F (0) = 0. Recall that° 1| oe dx = <J/T.0 2Explain using a modification of Problem 48 on Page 224 whyF'(b) = I ~2xe sin (2bx) dx + ape0F'(b) = 2b (/ e* cos (2bx) dx-+ er)1 bi=2b( F(b)+ VE VE2 2Now use the integrating factor method for solving linear differential equations frombeginning differential equations to solve the ordinary differential equation. If youhave not seen this method, it is just this. To solve y’ + f (x) y = g(x), multiply bothsides by e*) where F’ (x) = f (x). This reduces the left side to 4 (e" y), Thusa (er (0) = /72be~?”" Then do f db to both sides and use that F (0) = 4/7.) = 2bF (b) +/m2be”You can do the same problem as above using contour integration. For b > 0, use thecontour which goes from —a to a to a+ ib to —a+ib to —a. Then let a — © andshow that the integral of e~ over the vertical parts of this contour converge to 0.2_ p2 ..Hint: For z=x-+ib,e2 =e (+20) — ob” 6 (cos (2xb) + isin (2xb)).Consider the circle of radius | oriented counter clockwise. [,,z~ ® cos (z)dz =?The circle of radius | is oriented counter clockwise. Evaluate Sy Tcos (z) dz.Find fy” 2x dx.wo yt/3Find fy 24x.Suppose f is analytic on an open set U and a@ € U. Definee(2)= fof) itz Zof' (a) ifz=ashow that g is analytic on U.Let y* be aC! oriented closed curve and let a ¢ y* Then1 1n(Y, 0) = mai |e imaThis is called the winding number. Hint: Now let g(r) = f[ a ma edt. Show thatdoa Gr 8 (y(t) — )) = 0. Explain why this requires e~8 (y(t) — a) is a constant.dz=meENSince y parametrizes a closed curve, argue that g(b,) = 2mmi for an integer m.