9.11. EXERCISES 225

Now use the chain rule and the fundamental theorem of calculus to find f ′ (x) . Thenchange the variable in the formula for f ′ (x) to make it an integral from 0 to 1 andshow f ′ (x)+g′ (x) = 0.Now this shows f (x)+g(x) is a constant. Show the constantis π/4 by assigning x = 0. Next take a limit as x→∞ to obtain the following formula

for the improper integral,∫

∞

0 e−t2dt,(∫

∞

0 e−t2dt)2

= π/4. In passing to the limit inthe integral for g as x→ ∞ you need to justify why that integral converges to 0. Todo this, argue the integrand converges uniformly to 0 for t ∈ [0,1] and then explainwhy this gives convergence of the integral. Thus

∫∞

0 e−t2dt =

√π/2.

50. To show you the power of Stirling’s formula, find whether the series ∑∞n=1

n!en

nn con-verges. The ratio test falls flat but you can try it if you like. Now explain why, if n islarge enough, n!≥ 1

2√

π√

2e−nnn+(1/2) ≡ c√

2e−nnn+(1/2).

51. The Riemann integral∫ b

a f (x)dt for integrator function F (t) = t is only defined if fis bounded. This problem discusses why this is the case. From the definition of theRiemann integral, there is a δ > 0 such that if ∥P∥< δ , then the Riemann sum SP ( f )

must satisfy∣∣∣SP ( f )−

∫ ba f dt

∣∣∣< 1. Pick such a partition P = {a = x0 < · · ·< xn = b}and say SP ( f ) = ∑

ni=1 f (ti)(xi− xi−1) . Suppose that f is unbounded on

[x j−1,x j

].

Then you can modify the points ti, keeping all the same except for t j ∈[x j−1,x j

]and

let this one be t̂ j where this is chosen so large that

∣∣ f (t̂ j)(x j− x j−1

)∣∣−(∣∣∣∣∣∑i̸= jf (ti)(xi− xi−1)

∣∣∣∣∣+∣∣∣∣∫ b

af dt∣∣∣∣+1

)> 100

Show this is a contradiction. Hence f must be bounded.

52. Does the above conclusion that f is bounded hold in case of an arbitrary RiemannStieltjes integral assuming the integrator function F is strictly increasing?

53. Use Theorem 9.9.1 and Lemma 9.9.2 to justify the following argument. Let f becontinuous on [a,b]× [c,d] . Let

F (x)≡∫ x

a

∫ d

cf (t,y)dydt−

∫ d

c

∫ x

af (t,y)dtdy.

Then F is continuous on [a,b] . Also F (a) = 0 and

F ′ (x) =∫ d

cf (x,y)dy−

∫ d

cf (x,y)dy = 0

and so F (b) = 0 so Fubini’s theorem holds.

54. Let {an} be a sequence of positive numbers such that limn→∞ nan = 0 and for all n ∈N, nan ≥ (n+1)an+1. Show that if this is so, it follows that the series, ∑

∞k=1 an sinnx

converges uniformly on R. This is a variation of a very interesting problem found inApostol’s book, [3]. Hint: Use the Dirichlet partial summation formula on ∑kak

sinkxk

and show the partial sums of ∑sinkx

k are bounded independent of x. To do this,you might argue the maximum value of the partial sums of this series occur when∑

nk=1 coskx = 0. Sum this series by considering the real part of the geometric series,

∑qk=1

(eix)k and then show the partial sums of ∑

sinkxk are Riemann sums for a certain

finite integral.