Kenneth Kuttler

248 CHAPTER 10. IMPROPER INTEGRALS

Now differentiate and try to show that the derivative is negative for x ∈ (m,m+1).Thus the desired derivative is(

m−1

∑k=0

1x− k

− lnx

1Γ(x−m)

∫∞

0ln(t) tx−(m+1)e−tdt− 1

The first term is negative from the definition of ln(x) . The derivative being nega-tive will be shown if it is shown that the integral in the above is negative. Do anintegration by parts on this integral and split the integrals to obtain∫

∞

0ln(t) tx−(m+1)e−tdt = −

∫ 1

0tσ e−tdt +

∫ 1

0(t−σ)e−ttσ ln(t)dt

+∫

∞

1tσ e−t (1− (t−σ) ln(t))dt

where σ = (x−m)− 1 ∈ (−1,0) so −σ > 0. The last integral is negative because(t−σ) = t +(−σ)> 1. The first two integrals on the right are obviously negative.

17. Lemma 10.2.6 can be generalized. Suppose h, g are defined on all of R and that g isof bounded variation on every finite interval while h is continuous. Suppose also thatthere exists c, possibly not in [a,b] such that H (c)(g(b)−g(a)) =

∫ ba Hdg. Then

the same conclusion holds. That is∫ b

a ghdx = g(b)∫ b

c h(t)dt +g(a)∫ c

a h(t)dt. Thisgeneralizes the second mean value theorem for integrals.

18. Here is another version of Lemma 10.2.6. Let h be continuous and g is decreasingon [a,β ] and increasing on [β ,b] ,β ∈ [a,b]. Let these functions be defined on [a,b] .Then there exist c, ĉ in [a,β ] , [β ,b] respectively such that∫ b

aghdt = g(β )

∫ ĉ

ch(t)dt +g(a)

∫ c

ah(t)dt +g(b)

∫ b

ĉh(t)dt (10.12)

For the above identity to be valid, it is sufficient that c, ĉ∈ [a,b] satisfy the following.For H ′ = h

H (c)(g(β )−g(a)) =∫

aHdg,

H (ĉ)(g(b)−g(β )) =∫ b

Hdg (10.13)

24817.18.CHAPTER 10. IMPROPER INTEGRALSNow differentiate and try to show that the derivative is negative for x € (m,m-+1).Thus the desired derivative ism—1 1— —lInx acm Ih In(t) (m1) ot gy — —foo *—K T(x—m)The first term is negative from the definition of In(x). The derivative being nega-tive will be shown if it is shown that the integral in the above is negative. Do anintegration by parts on this integral and split the integrals to obtain[ In(t)e Med = -[ te ‘rh | t—o)e't? In(t) dt0 0+ fot ie (1—(t—6)In(1)) dtwhere o = (x—m)—1 € (—1,0) so —o > O. The last integral is negative because(to) =t+(-—o) > 1. The first two integrals on the right are obviously negative.Lemma 10.2.6 can be generalized. Suppose h, g are defined on all of R and that g isof bounded variation on every finite interval while 4 is continuous. Suppose also thatthere exists c, possibly not in [a,b] such that H (c) (g(b) — g(a)) = f° Hdg. Thenthe same conclusion holds. That is f? ghdx = g(b) 2 n(t)dt +g(a) [i h(t)dt. Thisgeneralizes the second mean value theorem for integrals.Here is another version of Lemma 10.2.6. Let be continuous and g is decreasingon [a, 8] and increasing on [Bb], B € [a,b]. Let these functions be defined on [a, 5].Then there exist c,é in [a, B] ,{B,b] respectively such that[erat =e p) [m0 )dt-+e(a ) [ne )dt-+9(b ) [near (10.12)For the above identity to be valid, it is sufficient that c, ¢ € [a, b] satisfy the following.For H’ =hH(é)(g(b)—g(B)) = }, Hdg (10.13)