Calculus

First of all, the usual rubbish:

∫ d
dx f (x ) d
dx f (x )
tan x sec2(x)
cot x –cosec2(x)
sec x ln (sec x + tan x )
ax ax ln(a)
1
1
a atan ( ax ) x 2 +a 2

2a ln ( x +a )
1 x −a
1
(x 2
> a 2
)
= − a1 acoth ( ax ) x 2 − a2
2a ln ( a −x )
1 a +x
1
(x 2 < a2)
= a1 atanh ( ax ) a − x
2 2

1
asin ( ax ) a 2 −x 2
1
asinh ( ax ) x +a 2
2

1
acosh ( ax ) x −a 2
2

Strategies…
Function of Strategy
[These also work if the top is a function of ax + b]
a2 − x 2
Substitute x = a sin θ
a −x
2 2

a2 + x 2 Substitute x = a tan θ
a2 + x 2 Substitute x = a sinh θ

x 2 − a2
x 2 − a2 Substitute x = a cosh θ
sech x Write in terms of exponentials and substitute u = ex
Rational function Substitute t = tan x2 , and then use the results
of sin x and/or 2t 1 −t2 2
sin x = cos x = dx = dt
cos x 1 + t2 1 + t2 1 + t2

1 − cos x Use the half-angle formulae to remove the root

Hold one factor of sin in reserve, and changes all
If m is odd the other sines to cosines. Then, substitute

x = cos θ
sinm x cosn x If n is odd Similar.
If neither are
Use the half-angle formulae to reduce the powers
odd
OR – expand into lots of sin nx using complex numbers!
Maths Revision Notes © Daniel Guetta, 2006
 1
(ax + b ) px 2 + qx + r
Substitute ax + b = 1
u

eax cos x Convert cos x into the exponential form of a complex number
x2 Re-write it as −1 + 1 + x 2
1 + x2 1 + x2

Partial fractions:
Fraction Decompose into
f (x ) A B
+
(x + α ) (x + β ) x +α x +β
f (x ) Ax + B C
+
(x 2 + α)(x + β ) x2 + α x +β
f (x ) A B C X
+ + + ⋯ +
(x + α) (x + β ) (x + β )
n n
x + α x + β (x + β )
2

• To prove the product rule:

o y (x ) = f (x ) g (x )
o y (x + δx ) = f (x + δx ) g (x + δx ) = [ f (x ) + δ f ][g(x ) + δg ] = f (x )g(x ) + g(x )δ f + f (x )δg + δgδ f
o δy y (x + δx ) − y(x ) δ f δg δg δ f
= = g(x ) + f (x ) +
δx δx δx δx δx
o δy δf δg δg δ f df dg
lim = lim g(x ) + lim f (x ) + lim = g(x ) + f (x )
δx →0 δx δx →0 δx δx →0 δx δx →0 δx dx dx
• To prove the chain rule, let δg be the fluctuation in g(x) as x increases by δx
δy δf δf  δ f δg  δf δg df dg
= = lim = lim   = lim × lim = ×
δx δ x δx → 0 δ x  δg δx  δx →0 δg δx →0 δx dg dx
δx →0 

• To prove integration by parts works:

d df dg
( fg ) = g + f
dx dx dx
dg d df
f = ( fg ) − g
dx dx dx
∫ fg ′ dx = fg − ∫ f ′g dx + C

• Leibnitz’s Formula is that

n  
dn dk dn −k
( fg ) = ∑  C k × k f × n−k
n
g
dx n 
k =0  dx dx 
To prove it:
N
o Assume true for N: f (N ) = ∑ NC k f (k )g (N −k )
k =0

o Differentiate with respect to x:

N
d (N ) d  (k ) (N −k ) 
f = ∑ NC k f g 
dx k =0 dx
N
f (N +1)
= ∑ NC k  f (k )g (N −k +1) + f (k +1)g (N −k ) 
k =0
N N +1
= ∑ Cs f g N (s ) (N −s +1)
+ ∑ NC s−1 f (s )g (N −s +1)
s =0 s =1

Maths Revision Notes © Daniel Guetta, 2006

 o Separate out the first term of the first series and the last term of the
other series:
N N
f (N +1)
= ( C 0 fg
N (N +1)
)+ ∑ N
Cs f g (s ) (N −s +1)
+ ∑ NC s −1 f (s )g (N −s +1) + ( NC N f (N +1)g )
s =1 s =1
N
= ( fg (N +1)
+f g ) + ∑ ( NC s + NC s−1 ) ( f (s )g (N −s +1) )
(N +1)

s =1
N +1
o Realise that NC s + NC s−1 = C s as follows:
N! N!
N
C s + NC s−1 = +
s ! (N − s ) ! (s − 1) ! (N − s + 1) !
(N − s + 1) N !+ sN !
=
s ! (N − s + 1) !
(N + 1) N ! (N + 1) ! N +1
= = = Cs
s ! (N − s + 1)! s ! (N + 1 − s ) !
o Simply feed it in
N
f (N +1)
= ( fg (N +1)
+f (N +1)
g ) + ∑ ( N +1C s ) ( f (s )g (N −s +1) )
s =1

o Realise that the first two terms are simply the thing inside the
summation at s = 0 and s = N + 1, and that therefore:
N +1
f (N +1)
= ∑ N +1C s f (s )g (N −s +1)
s =0

QED.

• Special points of a function:

df
o If = 0 , the point is a stationary point. In such a case:
dx
d2 f

If > 0 , the point is a minimum.
dx 2
2

If d f2 < 0 , the point is a maximum.
dx

Maxima and minima are also called turning points.
d2 f
o If 2
= 0 , we have a point of inflexion (whatever the value of df/dx).
dx

There will always be a point of inflexion between a maximum and
a minimum.
d2 f

If df/dx is also equal to 0, and 2
changes sign through the
dx
point, then we have a stationary point of inflection.
• Graph plotting:
o Find values as x gets very big and small, values at x = 0, and derivatives
of all these things.
o Mark the roots on the graph.
o Somehow show the envelope, if there is one.

Maths Revision Notes © Daniel Guetta, 2006

 o For graphs like ex cos x, make sure the period stays constant, and note
that maxima and minima aren’t as expected!
• Important simplifications:

A function is EVEN if f(x) = f(–x)

A function is ODD if f(x) = –f(–x)
a a
o For an EVEN function, ∫ −a
f (x )dx = 2∫ f (x )dx .
0
a
o For an ODD function, ∫ f (x )dx = 0 .
−a

o
∫ sinn (x ) = ∫ cosn (x ) = 0 as long as n is an ODD NUMBER.
whole number whole number
of periods of periods

• Stirling’s Formula:
n n
o First, we note that ln (n !) = ∑ ln (x ) ~ ∫ ln(x )dx for large n. Now,
1
x =1
n

∫ 1
ln(x )dx = n ln n − n + 1 ~ n ln n − n for large n. This gives

ln (n !) ≈ n ln n − n and n ! = n ne −n . Sadly, however, the latter is a bad

approximation, because a small eror in ln n! leads to a factor in n!
∞
o We note that the function Γ (n ) = ∫ x n−1e −x dx satisfies Γ (n + 1) = n Γ (n )
0

and Γ (1) = 1 . We therefore define n ! = Γ (n ) .

o Now, if we let x = n + y, then
y y2 y3
ln x = ln (n [1 + ]) = ln n + ln [1 + ] = ln n + − 2 + 3 + ⋯
y
n
y
n
n 2n 3n
∞
o Now, we can express Γ (n + 1) as Γ (n + 1) = n ! = ∫ e n ln x −x dx . Feeding the
0

expression we obtained for ln x into this, and integrating dy:

 
∞ y y 2
y 3  
n ! = ∫ exp n ln n + − 2 + 3 + ⋯ − (n + y ) dy

−n   n 2n 3n  
 
If n is sufficiently large, this is
 ∞ y y 2  

n ! = ∫ exp n ln n + − 2  − n − y  dy

−∞   n 2n  
 
∞  y2 
= ∫ exp n ln n + y − − n − y  dy
−∞  2n 
∞

∫
2
= e n ln n −ne −y / 2n
dy
−∞
∞

∫
2
n ln n −n
=e e −y / 2n
dy = n ne −n 2πn
−∞

• Differentiation of integrals:
b
o An integral ∫ a
f (x ) dx is a function of a and b. We can therefore
differentiate it with respect to either these two variables:

Maths Revision Notes © Daniel Guetta, 2006

 ∂ b


∂b ∫a
f (x ) dx = f (b) (since increasing the upper limit by δb will

increase the area by f (b)δb ).

∂ b


∂a ∫
a
f (x ) dx = −f (a ) (by swapping limits, or realising that

increasing the lower limit by δa will decrease the area by f (a )δa ).

b
o An integral ∫ a
f (x , λ) dx is a function of a and b and λ . We can therefore

differentiate it with respect to the parameter. It turns out that

∂ b b ∂
∂λ ∫a ∫
f (x , λ) dx = f (x , λ ) dx . This can be rationalised in one of two
a ∂λ
ways:

By noting that changing the parameter by δλ will change the
function everywhere in between the two limits, by an amount
∂
f (x , λ ) . We want the sum of all these changes over the
∂λ
function.

By using the definition of an integral as the limit of a sum:
∂ b ∂ ∂ b ∂
∂λ ∫a
f (x , λ ) d x =
∂λ
∑ f (x , λ ) δ x = ∑ ∂λ
f (x , λ ) δx = ∫a ∂λ
f (x , λ ) dx

o If a certain variable turns up both in the limits and as a parameter, then

just add contributions, ignoring the “corners”.
o This allows us to do a rather tricky integral. First, note that
∞
∞
∫ 0
e −ax dx = [− a1 e −ax ]0 = 1
a

We now differentiate this with respect to the parameter a:

∂ ∞ ∞ ∂ −ax ∞ ∂  1  1
∂a ∫ 0
e −ax dx = ∫0 ∂a
e dx = ∫
0
−xe −ax dx =  =− 2
∂a a  a
Doing this repeatedly, we find that
∞ n!
∫ 0
x ne −ax dx =
a n +1
Rather quicker than integration by parts! Note that if a = 1, this is the Γ
function.

• Schwatz’s Inequality is an extension of the triangle rule to inner products

(generalisations of dot products):

(∫ ) ≤∫
b 2 b b
fg dx f dx ∫ g 2 dx
2
a a a

To prove it:
b
o We know that ∫ ( f + λg ) dx ≥ 0 , because the function is positive
2

a
b b b
everywhere. So ∫ a
f 2 dx + 2λ ∫ fg dx + λ 2 ∫ g 2 dx ≥ 0 .
a a
b
o We can assume that ∫ a
f 2 dx ≠ 0 - otherwise, the both sides of the

inequality above are 0 and it’s trivially true.

Maths Revision Notes © Daniel Guetta, 2006

 o So, doing a bit of re-arrangement:
b b

1 + 2λ
∫ a
fg dx
+ λ2
∫a
g 2 dx
≥0
b b

∫ ∫
2 2
f dx f dx
a a

   b 
b 2 2 b
 ∫
1 + λ ab
fg dx   ∫ fg dx 
 − λ 2  a  + λ 2 ∫
a
g 2 dx
≥0
  
 f 2 dx 
 b b

∫a  ∫a ∫
2 2
 f dx   f dx
a

We now chose λ so that the first bracket becomes 0 (this involves

dividing by ∫ fg . If this is 0, Schwartz’s is trivially true):
 b 
2 b
 ∫ fg dx  ∫ g 2 dx
−λ2  ab  + λ 2 a
≥0
 f 2 dx   b
 ∫a ∫
2
 f dx
a

Since λ > 0, we can cancel and rearrange:

 b 
b 2

∫a  ∫ 
2
g dx  fg d x 
≥  ab 
b
 f 2 dx 
∫a  ∫a
2
f d x 
b

∫ g 2 dx
(∫ ) (∫ )
b 2 b 2
fg dx ≤ a
b f dx2

∫
a 2 a
f dx
a

(∫ )
b 2 b b
fg dx ≤ ∫ g 2 dx ∫ f 2 dx
a a a

• When doing simple multiple integrals in which the limits do not depend on each
other and the function can be separated into an x and a y component, a
simplification is possible:

(∫ )(∫ )
d b b d

∫ ∫
c a
g(x )h(y ) dx dy =
a
g(x )dx
c
h(y )dy

• The substitution c = cos θ is often very useful in spherical polars.

∞ ∞

∫ ∫
2 2
• We can use an elegant trick to evaluate e −x dx . We first let I = e −x dx
−∞ −∞
∞
and we note that since x is a dummy variable, we can also say I = ∫ e −y dy . So
2

−∞
∞ ∞ ∞ −(x +y
2 2
) −(x +y
2 2
)
∫ e −y dy ∫ ∫ ∫
2 2
I2 = e −x dx = e dx dy = e dx dy
−∞ −∞ −∞
2D Plane

We can now change to polar coordinates:

−(x 2 +y 2 ) ∞ 2π

∫ ∫ ∫
2
e dx dy = e −r r dr dφ = π
r =0 φ =0
2D Plane

Which means that

∞ ∞

∫ ∫
−x 2 2
e dx = π and e −x dx = 1
2 π
−∞ 0
2
Since e −x is an even function
Thus, the normalised normal distribution is given by
(x −µ)2
1 − (if X ~ N (µ, σ) )
P(X = x ) = e 2σ2
2πσ 2

Maths Revision Notes © Daniel Guetta, 2006