Kenneth Kuttler

136 CHAPTER 5. LINE INTEGRALS AND CURVES

17. Show k =√

2π as follows. Verify the following:∫π/2

0sin2m (x) =

(2m−1)(2m−3) · · ·12mm!

2∫ π/20 sin2m+1 (x) = 2mm!

(2m+1)(2m−1)···3 ,∫ π/2

0 sinn (x) = n−1n∫ π/2

0 sinn−2 (x) . Show the fol-lowing using the above.

(2m+1

)1≥

∫ π/20 sin2m (x)

2m2m+1

∫ π/20 sin2m−1 (x)

≥1∫ π/20 sin2m (x)∫ π/2

0 sin2m+1 (x)

=(2m−1)(2m−3)···1

2mm!π

22mm!

(2m+1)(2m−1)···3=

(2m+1)(2m−1)2 (2m−3)2 · · ·12

4m (m!)2π

2≥ 1

Then multiply on the top and bottom by (2m)2 (2(m−1))2 · · ·22. Obtain

limm→∞

2mm!2m (m!)√(2m+1)(2m)!

√2π= 1

Now limm→∞(m!)2e2m

k2m2m+1 = 1 = limm→∞(2m)!e2m

k(2m)2m+(1/2) where k is the constant of the above

problem. Thus

limm→∞

k2m2m+1 (2m)!e2m

(m!)2 e2mk (2m)2m+(1/2) = 1

Now

limm→∞

2mm!2m (m!)√(2m+1)(2m)!

√2π

k2m2m+1 (2m)!e2m

(m!)2 e2mk (2m)2m+(1/2)

= limm→∞

1√(m+(1/2))

√1

2πkm1/2 =

√1

2πk = 1

136 CHAPTER 5. LINE INTEGRALS AND CURVES17. Show k = 27 as follows. Verify the following:m/2 (2m—1)(2m—3)---122m _[ sin“” (x) = am | 5hi * Sint] (x) = Cmts? ha * sin” (x) = nt 7/2 sin”2 (x) Show the fol-lowing using the above.>1& ') Jo sin?" (x)? sin” (x)2m ~ wy a * Sinz! (x) I’ ? intl (x)Cm (2m+ 1) 2m=1)? Qm—3)?- 1?2m! m 2(2m+1)(2m—1)-3 4” (m!)>1NIAThen multiply on the top and bottom by (2m)? (2 (m—1))*-++2?. Obtain2"ml2"(m!) [2—————m+ \/(2m-+1) (2m)! V: (m!)?e2” 4-1: (2m) !e2”" :Now limo amt = 1 = lim, 500 iam" where k is the constant of the aboveproblem. Thuslam kent! (2m) 1e2mm—yco (m!)* e2"k (2m)?"*/2) =1Now2”m!2" (m!) 2 km?! (2m)!e2"imme \/(2im +1) (2m)! VM (ml)? ek (2m?)— im —— J et? =, J 2 pK= km, mum Wak !