1. Let c ∈ R , and define a function  f : [−1, 1] → R  by

f(x) =  北(1) 

Is the function  f  integrable on the interval  [−1, 1] ?

Solution:

Yes,  f  is integrable on the interval  [−1, 1] .  Clearly,  f  is continuous on  [−1, 0) ∪ (0, 1] , and only possibly discontinuous at the point 0 . Since  f  is bounded ( |f(x)| ≤ max{1, |c|} for all x ∈ [−1, 1] ), by using a proposition of the lecture, we infer f  is integrable as desired. □

2. Let  ⌊x⌋  denote the integral part function. Is the function    f(x) := {1北0, − ⌊ 北(1)

integrable over the interval  [0, 1] ? (Hint: draw a graph of the function  f .)

Solution:

Yes,  f  is integrable over the interval  [0, 1] .   Since  f  is bounded  ( |f(x)|  ≤  1  for all x ∈  [−1, 1] ), by a lemma of the lecture we only need to show for every  6 ∈  (0, 1) ,  f  is integrable on [6,1] . Let us fix 6 ∈ (0, 1) , and the function f has only  ⌊ ⌋−1 discontinuous points on  [6,1] . So, we may apply a proposition of the lecture to see that  f  is integrable

on  [6,1] . This implies  f  is integrable over the interval  [0, 1] as desired.                       □

3.  Compute the integral  '4(北) 北   with the substitution t = tan x . Solution:

If t = tanx , then we have 1+t2  = 1+tan2 x = cos(1)2 北  . So, cos2 x =   and sin2 x =  . As  (tanx)\  = cos(1)2 北  , we get

\  = \1 ∞   · dt = \1 ∞ dt = [ −  − t−1]1(∞)  =  .

4.   (a) Prove by induction that the following identity

\− cos2n tdt = π (  )

holds for all n ∈ N+. Here,

u    2i    := 2 · 4 · 6 · ... ·     2n   .

Solution:

Let us prove it by induction on n .

Induction basis: When n = 1 , we have

\− cos2 tdt = \−  dt = [  + ]  =  = π ·  ,

as desired. Actually, the statement also holds for n = 1 if ni(0)=1 ci  := 1  (which is the standard definition of the product over an empty set).  Hence, we could also choose n = 0 as base case, which would be much easier to prove.

Induction hypothesis: Suppose the identity is true for n = k , for some  k ∈ N+ .         Induction step: We prove the identity for  n = k + 1 . Using integration by parts, we see that, for all ℓ ∈ N+ ,

π                                                       π

\− π(2)  cos2ℓ tdt = \− π(2)  (cos2ℓ−1 t)(sint)\ dt

2                                                      2

π

=[(cos2ℓ−1 t)sint] − \− π(2)  (sint) · (2ℓ − 1) · (cos2ℓ−2 t) · (−sint)dt

2

π

=(2ℓ − 1) \− π(2)  (1 − cos2 t)cos2ℓ−2 tdt

2

=(2ℓ − 1) (\− cos2ℓ−2 tdt) − (2ℓ − 1) \− cos2ℓ tdt.

Therefore, the identity

\− π(2)  cos2ℓ tdt =  \− π(2)  cos2ℓ−2 tdt

holds for all ℓ ∈ N+ . Consequently, in combination with the Induction hypothesis, we have

\− cos2(k+1) tdt =  \− cos2k tdt =  · π (  ) = π (  )

and this proves the identity when n = k + 1 .

Therefore, the identity holds for all n ∈ N+ .                                                           □

(b) Compute the integral  \   for every positive integer  n  with the substitution x = tant .

Solution:

If x = tant , then we have  1 + x2  = 1 + tan2 t =  . As   =  , we get

\−   = \− cos2n t · dt = \− cos2(n− 1) tdt.

By using part (a), we have

\ ∞       dx     

= 1

if n = 1

) ,   if n ∈ N+ / {1} .

□

5. Prove that if the integral \ f(x)dx is convergent, then we have limn→∞ \1 f(nx)dx = 0 .

Solution:

By using the substitution t = nx , we get

\−1(1) f(nx)dx = \−n(n) f(t) ·   =  .

As  \ f(x)dx  is convergent, we have

n(l) \−n(n) f(t)dt = c,

for some c ∈ R  (think about it!). Therefore, we infer

n(l) \−1(1) f(nx)dx =n(l)   = 0,

as desired.                                                                                                                     □

6.  Is the calculation

\−1(1) dx = [ln|x|]1  = 0.

correct?

Solution:

No, it is not correct. Indeed, the function    has a singularity at the point  x = 0  and we may try it as follows

\−1(1) dx = \0 1 dx + \−1(0) dx = (6l1+  \611 dx) + (6l2− \ dx)

= (6l1+  [ln|x|]6(1)1 ) + 6l2−  [ln|x|]1  = ∞ + (−∞).