Factorials – solutions.

1. (a) Prove that  e  x x N dx  N ! for any natural N.
0
1
(b) Prove that  x K 1  x  dx 
M
0
K !M !
for natural K, M.
 K  M  1!
Solution. (a) Induction. For N=0 You get integral = 0.


The step of induction: proving that  e x dx  N  e  x x N 1dx .
x
N
0
0
N
Apply integration by parts. First function is x , the derivative of the second
is e  x . Minus (of deriving e  x ) times minus (of integration by parts) is plus.
(b) I think that the first solution (I saw it in Euler’s book) was integration by
M
parts (integrate x K , differentiate 1  x  ).
However, there is another classical solution I like better.
Consider double integral over quadrant (¼ plane):

I ( K , M )   x K y M e  x  y dxdy
0 0
We shall compute it in two ways.
First way: split it into product of two integrals.

I ( K , M )   x y e
K
M
0 0
 x y

0
Second way: substitute z  x  y, t 

K

dxdy   x e dx  y M e  y dy K !M !
M
K x
0
x
which is x  zt, y  z  zt
x y

M
 x  y
I ( K , M )       z K  M e x y dxdy  t K 1  t  z K  M e z dxdy
z
z
0 0  
0 0
To continue this computation we need to convert dxdy into dzdt which is
done by the Jacobian.
x 
 x
z  t z 

z
t   t
J


 z
 y
y  1  t  z  1 0 
t 
 z
So dxdy = |J|dzdt = zdzdt, hence

I ( K , M )   t 1  t  z
M
K
1
e dxdy   t K 1  t  z K  M 1e  z dzdt 
K M  z
M
0 0

 z
0 0
1
1
e dz  t 1  t  dt   K  M  1!   t K 1  t  dt
K  M 1  z
0
K
M
M
0
0
The two answers for the same questions that we have computed in two ways
1
should be equal, hence  K  M  1!   t K 1  t  dt  K !M !
M
0
QED.
Remark. First part of the question gives a natural extension of N! to all real
non-negative numbers, or in fact to all complex numbers for which ReN ≥ 0.
In fact, it can be extended to all complex plane, and it will have poles
(infinite values) at all negative integers. The standard notation is

  N    e  x x N 1dx   N  1!
0
and it is called “gamma function” or “Euler’s gamma function”.
Different mathematicians have found some nice properties of  and proved
several theorems of the type “why  is the most natural extension of
 N  1! to the positive/complex numbers”.
Here are some colored pictures http://mathworld.wolfram.com/GammaFunction.html
As for the second part, it allows to build the natural extension of binomial
coefficients to non-integer number. The integral we asked about is usually
denoted B(K+1,M+1) and called “beta function”.
 pq
The formula we proved in standard notation looks so:   p, q  
.
 p  q
2. Define (using 1a) and compute ½!
Solution. As it was explained in the above remark, the natural definition is
 

1 !  e  x xdx . I don’t believe there is an elementary function, whose

2
0
derivative is e x x . So, we need a trick.
Substitute K = M = ½ in the formula 1(b):
1
2!   t 1  tdt 
0
  
1 !
2
2
1
So. It is enough to compute
 t 1  t dt . This integral is related to a circle.
0
1
1
 1
 1
Really, let u  t  . Then t  t  1    u   u    u 2 and
2
2
 2
 4
1

0
1
t 1  t dt 
2

1
1
 u 2 dt .
4
2
This is precisely the are between diameter of a circle of radius ½ and its arc,
which is half a circle of radius ½ so it is  .
8
2
So, we have 2!   8  1 2 ! and hence 1 !  2!      
2
8
4
2
  
 
Remark. The calculator in Windows knows it 
3. Compute the volume of the N-dimensional ball of radius 1 and the area of
its surface.
Solution. Denote SN the area of its surface, VN the volume. Firstly,
SN = NVN
This easy fact can be explained in 2 ways.
First explanation. Divide the surface into small (or infinitesimal) countries.
Divide the volume into conic parts, whose vertex is the center of the ball,
and bases are countries on the surface. Each N-dimensional cone can be
computed as Sh/N. In our case h=1, so each volume part is N times smaller
than corresponding area part. This constant ratio will be preserved after
integration.
Second explanation. To compute volume of the ball, we shall split it into
thin concentric spherical layers (like cabbages). At radius R we have layer of
width dR and area SNRN-1 , and integration of this gives
1
S
VN   S N R N1dR  N .
N
0
So, it is sufficient to compute the area or the volume and not both.
First solution. Let z be one of the coordinates. The hyperplane at level z
intersects the ball along a thin layer of radius 1  z 2 , hence its N  1dimensional volume is VN 1 


1  z2
integral of this expression times dz.
1
VN   VN 1 
1
N 1

. The total volume is given by the
1 z

2
N 1
dz
This integral begs for trigonometric substitution:
z  cos x
dz   sin xdx
1  z 2  sin x
Then we get

VN  VN 1   sin x  dx
N

So now we have to solve 4(a) and multiply the answers.
The details are left to the reader (if he or she will really want to finish the
first solution after they see the second one).
Second solution. Consider N-dimensional integral:
 
GN 


 ...  e
 

 x12  x22 ... xn2

dx1dx2 ...dxN

We shall compute it in 2 ways. First way is splitting it into product:

GN 
e
 x12

dx1 

e

 x22

dx2  ... 
x
N
 e n dxN  G1
2

Second way is integrating it thing cabbage-like spherical slices of radius R
centered at zero. The volume of each slice is Sn R N 1dR so we get

GN   e  R S n R N 1dR , now substitute u  R2 , du  2RdR :
2
0


 
Sn  R2 N 2
Sn  u N2 1
S
N
 S
GN   e R 2 RdR   e u du  n  N    1! n
2
2 0
2 0
2
2
 2
N
 S
N
So, G1N    1! n and G1N   ! Vn . To conclude this solution, we have
2
 2
2
yet to find G1.
Instead of computing it directly, use the fact that we know the area of the
unit circle in plane. G   . The answer: VN 
2
1


N
N
2
, SN  N
!
2
 2 ! or  3 2 ! we say that


N
N
2
!
2
.
Remark. And if someone asks what is 1
  12 !  and  N  1! N! N  1

1 3
hence  1 ! 
,  3 !     ,  5 ! 
2
2
2
2
2 2
1 3 5
2 2 2
Of course, for N=2k+1 we can write the formula in usual notations
VN  2
 2 
 N 1
   .
 N 1
2
,
SN  2
 2 
2
,
1  3  ...  N
1  3  ...   N  2 
where the denominators contain products of first odd numbers.

4*. (a) Let I N   sin N  x  dx . Compute I N for each integer N.
0
(b) Show that
 2N 
1

N  4 N 
  N  

2

N 
Solution. (a) Use integration by parts. sin N  x  is a product of 2 functions:
first , sin N 1  x  , second sin  x  . We will differentiate the first and integrate
the second. The integral of sin is - cos, but minus will cancel out with minus
of integration by parts. The value at the limits is 0, hence:

I N   sin
N 1
0

 x  sin  x  dx   N  1  sin N 2  x  cos  x  cos  x  dx 
0

  N  1  sin N 2  x  1  sin 2  x   dx   N  1  I N 2  I N 
0
Hence I N   N  1  I N 2  I N  so NI N   N  1 I N 2 or I N 
N 1
I N 2 .
N
It is easily computed that I 0   and I1  2 .
2k  1 2k  3
1
2k 2k  2

    , and I 2 K 1 


So, I 2 K 
2k 2k  2
2
2k  1 2k  1
2
 2.
3
Remark. To write those formulas shortly, special notations were invented.
The product of all natural numbers up to N having the same parity as N is
 2k  1!!   and I   2k !!  2 .
denoted N!! . In this notation, I 2 K 
2 K 1
 2k !!
 2k  1!!
(b) The value of sin is not bigger (and usually smaller) than 1.
Hence the value of sin N  x  decreases as N grows, at least on the interval
0,  . That is why I N
is decreasing sequence. So, we can write
I 2 N  I 2 N 1  I 2 N 2 .
One obvious consequence of this is so-called Wallis formula.
Since I 2 N  I 2 N  2 , and is I 2 N 1 between, I 2 N  I 2 N 1 hence

 2 N !!
2
2 2 4 4 6 6
      
2  2 N  1!! 2 N  1!! 1 3 3 5 5 7
But we shall pay more attention and obtain more precise inequalities.
First we shall reduce !! to ! .
 2 N !!  2  4  6   2 N  2 N 1  2  3   N  2 N  N !
 2 N  1!! 

 2 N  1!   2 N  1!
 2 N !! 2 N  N !
Now lets develop I 2 N  I 2 N 1  I 2 N 2 :
 2 N  1!!  2   2 N !!  2   2 N  1!!  2
 2 N !!
 2 N  1!!  2 N  2 !!
 2 N  1 


  2 N !!
 2 N  1  

2
2
  2 N  1!! 2 N  2 2
  2 N !!  4 N  4 N  
1

N






2
2N  2 2

  2 N !
2
2
4
2
2
  N ! 2 N 2
1

 N      
2
 2 N !


1
4N

 N      2N !
 
2

2

  N 


 N !
 N 
2
 N N! 
5. Compute lim 
.
N 
N


First solution.
Lemma 1. If sequence {an} has a limit, then the sequence of its averages
 a1  a2  ...  an 

 has the same limit.
n


Proof is direct, with epsilons and deltas. All element of the second sequence
after some index are near the limit, and others give finite (and reducing)
contribution.
Lemma 2. To find a limit of a sequence {bn} it is enough to find the limit of
the sequence {(n+1)bn+1 – nbn}, and if we do, the limit of {bn} is the same.
Proof : it follows directly from lemma 1, since sequence {bn} are averages of
sequence {(n+1)bn+1 – nbn}.
Lemma 1’. If sequence of positive numbers {an} has a limit, then the
sequence of its geometric means

n

a1a2  ...  an has the same limit.
Lemma 2’. To find a limit of a sequence of positive numbers {bn} it is
 b n 1 
enough to find the limit of the sequence  n n1  has the same limit.
 bn 
The last two lemmas follow from the former two by taking exp of all
sequences. We shall use lemma 2’, but I wanted to show you other things
which are useful.
N
N!
So, for limit of
it would be enough to find the limit of
N
 N  1! N !  N N  1
N 1
N
N N  N  1 N 
1
 N  1
1  
N

The denominator is a famous expression for e, so this sequence and the
original one two converge to 1 .
e
Second solution (I learned it from a 13-year old who invented it himself, Gil
1
2
N
 N N !  ln N  ln N  ...  ln N
Gitik). Take the log. ln 

N
N


1
This is Riemann sum for an integral  ln xdx   x ln x  x 0  1 .
1
0
1
So, before taking ln, the limit was e .