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Copyright (C) 2013 Frédéric Wang
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If is a primitive of then ,
provided the integrand and integral have no singularities on the path of
integration. As a consequence, we can use a table of derivatives:
|Linearity || || |
|Leibniz rule || || |
|Reciprocal rule || || |
|Chain Rule || || |
|Inverse function rule || || |
|Elementary power rule || || |
|Generalized power rule || || |
| Exponential || || |
| Logarithm || || |
| Sine || || |
| Cosine || || |
| Tangent || || |
Using linearity and elementary power rule:
Using linearity and sine/cosine:
Using Leibniz rule, Chain rule and Exponential/Power:
Using the inverse function rule and tangent:
Using linearity, reciprocal rule and logarithm:
Using the power rules and chain rule:
Integration by Substitution
Do a substitution to simplify an integral. We have
: this is the chain rule!
Let , then we get
We find the same result as
Let , so
Let , :
Let and so . We have
We have on
We first write
Now doing the substitution ,
But we have already met this integral in the basic methods:
We do the substitution . We have
. Also, we have
Integration by parts
Given a two function we have
this is Leibniz rule!
We consider and that is
and . The integral
can be written and hence we get
We now consider the second integral can be written
We now recognize in the last integral and so
We let and . Hence
and . The integration
by parts gives:
The integral can be written
We do an integration by parts with and
We repeat a similar integration by parts on the remaining integral:
We note that we found the initial integral . We get:
Integration by Partial Fraction Decomposition
Given a rational function , write its partial fraction
decomposition and integrate each term.
Clearly, we have and is a trivial root of the second factor so we can
factor it by and obtain:
Hence we search a decomposition of
for some constants . We have
The denominator can be written
. Evaluating the
second factor at we don't find any trivial roots. Hence we try
to convert it to a depressed quartic by setting .
We find .
The discriminant of is so we have the decomposition
in irreducible factors:
We now try to find such that
From the two last equalities, we immediately get and . Finally,
By polynomial long division we find that
The integral of the first term is:
First we search trivial roots for the denominator.
If we evaluate at we find respectively
. So we can factor by . We get
The polynomial is irreducible on because its
discrimant is . Hence we search a decomposition of
for some constants . We have
The integral of the two first terms is easy to compute:
The last term can be split into two parts:
Finally we get
Alternatively, we can write
where are the complex roots of
. We have
After simplification, we get
as we already found above.
Consider contour integral along paths in complex
plane to deduce integral along the real line.
Consider the integral
Write , . The integral becomes
where is the unit circle traversed counterclockwise.
singularities inside that circle: and .
Hence the integral is the times the sum
of the residues of at these points:
We want to know the value of the integral
For that purpose, we introduce the function
We note that is the integrand above. We also note that
poles of are . By choosing the right contour, one can