Integration of Complex Functions

Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function ƒ.The integration process of complex functions involves the operation of doing integration with the complex functions.(source from Wikipedia)


Examples to Explain "Application Integration of Complex Functions "

Integrate the function involving  complex function numbers:  int e^x sin(x) dx

Solution:

By rearranging

int e^((1+i)x) dx  =  int e^x . e^(i.x) dx  =  int e^x . [cos(x) +i sin(x)] dx

=   int e^x cos(x) dx + i int e^xsin(x) dx

Solve int e^((1+i)x) dx

Let u =  (1 + i) x then

du = (1 + i) dx      =>         dx =  (du)/(1+i)

On substituting we get,

1/(1+i) int e^u du    =>          e^u/(1+i)       =>       e^((1+i)x)/(1+i)

Multiply with (1-i)/(1-i)     to remove the deominator of complex number

e^((1+i)x)/(1+i)      (1-i)/(1-i)      =>        (1-i)e^((1+i)x)/(1^2-i^2)       =>        (1-i)e^((1+i)x)/(1-(-1))        =>        (1-i)e^((1+i)x)/2

Here concentrate in the integral constant

int e^((1+i)x) dx  =     (1-i)e^((1+i)x)/2     + C1 + i C2

By rearranging

int e^((1+i)x) dx  =   (1-i)e^((1+i)x)/2   + C1 + i C2  and applying formula generated by Euler.

=     (1/2-i/2)e^x[cos(x) + i.sin(x)]   + C1 + i C2

=    1/2e^x[cos(x) + i.sin(x)]   -i/2e^x[cos(x) + i.sin(x)]   + C1 + i C2

=    1/2e^x[cos(x) + i.sin(x)]   -1/2e^x[icos(x) - sin(x)]   + C1 + i C2

=      1/2e^xcos(x)  +   1/2e^x i.sin(x) -1/2e^x icos(x)  +1/2e^x sin(x)  + C1 + i C2

=     1/2e^x[cos(x) + i.sin(x)]  + C1 -i/2e^x[sin(x) - cos(x)]  + i C2

=     1/2e^x[cos(x) + i.sin(x)]  + C1  + i.(1/2e^x[sin(x) - cos(x)]+C_2

Since the imaginary part is concerned with the sin x then the result only in the imaginary values and neglect the real values.

int e^x sin(x) dx       =    1/2e^x[sin(x) - cos(x)]+C_2    is the required integration result.

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Problems to Explain "integration of Complex Functions "

Integrate the function with complex functions:  int ie^((5+i)x) dx

Solution:

Let u =  (5 + i) x then

du = (5 + i) dx      =>         dx =  (du)/(5+i)

On substituting it in the above function we get,

i/(5+i) int e^u du    =>          e^u/(5+i)       =>       ie^((5+i)x)/(5+i)

Multiply with (5-i)/(5-i)     to remove the deominator of complex number

ie^((5+i)x)/(5+i)      (5-i)/(5-i)      =>        i(5-i)e^((5+i)x)/(5^2-i^2)       =>        i(5-i)e^((5+i)x)/(25-(-1))        =>        (5i-i^2)e^((5+i)x)/26

Here concentrate in the integral constant

int ie^((5+i)x) dx  =   (5i+1)e^((5+i)x)/26   + C1 + i C2