224 CHAPTER 9. INTEGRATION

This is called the second mean value theorem for integrals. Hint: Use integration byparts. ∫ b

agd f =−

∫ b

af dg+ f (b)g(b)− f (a)g(a)

Now use the first mean value theorem, the result of Theorem 9.9.1 to substitute some-thing for

∫ ba f dg and then simplify.

46. Generalize the result of Theorem 9.9.4 to the situation where α and β are only ofbounded variation.

47. This problem is in Apostol [2]. Explain why whenever f is continuous on [a,b]

limn→∞

b−an

n

∑k=1

f(

a+ k(

b−an

))=∫ b

af dx.

Apply this to f (x) = 11+x2 on the interval [0,1] to obtain the very interesting formula

π

4 = limn→∞ ∑nk=1

nn2+k2 .

48. Suppose f : [a,b]×(c,d)→R is continuous. Recall the meaning of the partial deriva-tive from calculus,

∂ f∂x

(t,x)≡ limh→0

f (t,x+h)− f (t,x)h

Suppose also ∂ f∂x (t,x) exists and for some K independent of t,∣∣∣∣∂ f

∂x(t,z)− ∂ f

∂x(t,x)

∣∣∣∣< K |z− x| .

This last condition happens, for example if ∂ 2 f (t,x)∂x2 is uniformly bounded on [a,b]×

(c,d) . (Why?) Define F (x) ≡∫ b

a f (t,x)dt. Take the difference quotient of F andshow using the mean value theorem that F ′ (x) =

∫ ba

∂ f (t,x)∂x dt. Is there a version of

this result with dt replaced with dα where α is an increasing function? How aboutα a function of bounded variation?

49. I found this problem in Apostol’s book [2]. This is a very important result andis obtained very simply by differentiating under an integral. Read it and fill in

any missing details. Let g(x) ≡∫ 1

0e−x2(1+t2)

1+t2 dt and f (x) ≡(∫ x

0 e−t2dt)2

. Note

∂

∂x

(e−x2(1+t2)

1+t2

)=−2xe−x2(1+t2) and

∂ 2

∂x2

(e−x2(1+t2)

1+ t2

)=−2e−x2(1+t2) +4x2e−x2(1+t2) +4x2e−x2(1+t2)t2

which is bounded for t ∈ [0,1] and x ∈ (−∞,∞) . Explain why this is so. Also showthe conditions of Problem 48 are satisfied so that

g′ (x) =∫ 1

0

(−2xe−x2(1+t2)

)dt.