4 Logarithm and exponential functions
4.1 Logarithm
We have seen that, for
a
=
−
1,
x
a
dx
=
x
a
+1
a
+ 1
a
=
−
1
,
and we still have to define this integral for
a
=
−
1. For this, we introduce the logarithmasln
x
=
x
1
duu,
so that the logarithm gives the surface area between the function 1
/u
and the horizontalaxis, from 1 to
x >
0. The real logarithm is not defined for
x <
0, since the correspondingsurface area would be infinite. The number
e
is defined by ln
e
= 1 and
e
≃
2
.
718281828.
Properties
:
•
We have seen that the integrals
∞
1
dx/x
and
10
dx/x
both diverge, such thatlim
x
→∞
ln
x
= +
∞
and lim
x
→
0
ln
x
=
−∞•
From the definition of the logarithm, one can see thatln(
ab
) =
ab
1
duu
=
a
1
duu
+
aba
duu
= ln
a
+
b
1
dvv
= ln
a
+ ln
b,
(6)where we make the change of variable
u
=
av
.
•
One can also see thatln(
x
a
) =
x
a
1
duu
=
x
1
advv
=
a
ln
x,
(7)where we make the change of variable
u
=
v
a
.
•
We have in general, for any differentiable function
f
,
dxf
′
(
x
)
f
(
x
)= ln
|
f
(
x
)
|
+
c,
where
c
is a constant18
Logarithm in base
a
The logarithm in base
a
is defined aslog
a
(
x
) =ln
x
ln
a,
and is equal to 1 when
x
=
a
. Note that ln
x
= log
e
(
x
).
Integral of the logarithm
To calculate
ln
x dx
, one uses an integration by parts:
ln
x dx
=
(
x
)
′
ln
x dx
=
x
ln
x
−
x
(ln
x
)
′
dx
=
x
ln
x
−
x
+
c,
where
c
is a constant.
Limits
•
When
x
→
+
∞
:
We show here the important limitlim
x
→
+
∞
ln
xx
a
= 0
, a >
0 (8)which means that any (positive-) power law goes quicker to infinity than the loga-rithm, when
x
→
+
∞
.
Proof
For any
u
≥
1 and for any
a >
0, we have1
u
≤
1
u
1
−
a/
2
.
Integrating this inequality from 1 to
x
leads to0
<
ln
x
≤
2
a
(
x
a/
2
−
1)
<
2
ax
a/
2
.
Dividing by
x
a
gives the expected result:0
<
ln
xx
a
≤
2
ax
−
a/
2
→
0 when
x
→
+
∞
.
•
When
x
→
0
:
Another important limit to know islim
x
→
0
x
a
ln
x
= 0
, a >
0
,
(9)which means that any (positive-) power law kills the divergence of the logarithm at
x
= 0.
Proof
For any
u
satisfying 0
< u
≤
1 and any
a >
0, we have1
u
≤
1
u
1+
a/
2
.
Integrating this inequality from
x
to 1, we obtain0
<
−
ln
x
≤
2
a
(
x
−
a/
2
−
1)
<
2
ax
−
a/
2
.
Multiplying by
x
a
gives the expected result:0
≤
x
a
|
ln
x
|≤
2
ax
a/
2
→
0 when
x
→
0
.
19
4.2 Exponential
The exponential is defined as the inverse function of the logarithm:
y
= ln
x
⇐⇒
x
= exp
y
=
e
y
From property (6), if we note
u
= ln
a
and
v
= ln
b
, we haveexp(
u
+
v
) = (exp
u
)
×
(exp
v
)
,
and from property (7), if we note
y
= ln
x
, we have
exp
y
a
= exp(
ay
)
.
Derivative of the exponential
one can differentiate the definition exp(ln
x
) =
x
, which,using the chain rule, leads to exp
′
(ln
x
) =
x
. We therefore conclude that the derivative of the exponential is the exponential itself:exp
′
x
= exp
x.
Exponential of base
a
This function is defined as
a
x
= exp(
x
ln
a
)
,
which is consistent with the properties of the logarithm and the exponential. It’s derivativeis then(
a
x
)
′
= (ln
a
)
a
x
.
One can also define the function
x
x
, with derivative(
x
x
)
′
=
ddx
exp(
x
ln
x
)
= (1 + ln
x
)
x
x
.
Limits
•
From the limit (8), if we note
y
= ln
x
and
b
= 1
/a >
0, we havelim
y
→
+
∞
exp
yy
b
= +
∞
,
and the exponential goes to infinity quicker than any power law.
•
From the limit (9), if we note
y
=
|
ln
x
|
and
b
= 1
/a >
0, we havelim
y
→
+
∞
y
b
exp(
−
y
) = 0
,
and the decreasing exponential kills the divergence of any power law.20
4.3 Hyperbolic functions
The hyperbolic functions are defined as
•
hyperbolic cosine: cosh
x
= (
e
x
+
e
−
x
)
/
2;
•
hyperbolic sine: sinh
x
= (
e
x
−
e
−
x
)
/
2;
•
hyperbolic tangent: tanh
x
= sinh
x/
cosh
x
;
•
hyperbolic cotangent: coth
x
= cosh
x/
sinh
x
,and their derivatives are given bycosh
′
x
= sinh
x
sinh
′
x
= cosh
x
tanh
′
x
= 1
−
tanh
2
x
coth
′
x
= 1
−
coth
2
x
It can easily be seen that, from their definition, the functions cosh and sinh satisfy, for all
x
,cosh
2
x
−
sinh
2
x
= 1
.
Also, it can be easily checked thatcosh(2
x
) = cosh
2
(
x
) + sinh
2
(
x
)sinh(2
x
) = 2sinh(
x
)cosh(
x
)21
