# B Complex Integration By Partial Fractions

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1. Apr 8, 2016

### Brandon Trabucco

Hello, I am enrolled in calculus 2. Just having started a section in our text book about integration by partial fractions, I eagerly began trying to use this integration technique wherever I could. After messing around for multiple days, I ran into this problem:

∫ 1/(x^2+1)dx

I immediately wanted to split the denominator into a complex difference of squares and rewrite it as a sum of two fractions, just to see what would happen if I did so. I took the following steps to complete the transformation:

1/((x + i)(x - i)) = A/(x + i) + B/(x - i)

1 = A(x - i) + B(x + i)

1 = Ax - Ai + Bx + Bi

therefore we can assume the following:

A + B = 0

A = -B

Bi - Ai = 1

Bi + Bi = 1

B = 1/2i

A = -1/2i

After inserting these values into our partial fraction, I was left with the following equation:

1/((x + i)(x - i)) = -1/(2i(x + i)) + 1/(2i(x - i))

Then I took the integral of both sides, which should both be equivalent expressions, and simplified:

∫ 1/((x + i)(x - i))dx = -1/2i∫ 1/(x + i)dx + 1/2i∫ 1/(x - i)dx

arctan(x) = -(1/2i)ln(x + i) + (1/2i)ln(x - i)

arctan(x) = (1/2i)ln((x - i)/(x + i))

(2i)arctan(x) = ln((x - i)/(x + i))

After this step, my gut told me to exponentiation both sides to remove the natural logarithm. I question doing this however, because I am not familiar with the behavior with the natural logarithm in the complex plane, and I do not know if exponentiation will alter the behavior of this equality. Despite this, I set both sides as a power of e and ended up with:

e^((2i)arctan(x)) = (x - i)/(x + i)

This equation slightly resembles Euler's identity. Even more so, When taking the limit of both sides as x approaches positive infinity, the equation turns into the following:

e^(iπ) = 1 + 0i

This result immediately threw red flags in my head since I knew that e^(iπ) equals negative one. I worked through this problem multiple times and ended with the same result each time. My question for all of you would be whether or not my assumption of using the exponential to cancel the natural log works with complex numbers. If there is any other aspect of this proof that I have misinterpreted, then I would appreciate feedback to help me better understand the nature of this problem.

*I have made a note to myself that the complex graphs of e^((2i)arctan(x)) and (x - i)/(x + i) are in fact reflections of one another over the real axis; however, I have checked by algebra and integration steps and cannot find where the missing negative was from.

Thank you,
Brandon Trabucco

Last edited: Apr 8, 2016
2. Apr 9, 2016

### PeroK

Just three words: constant of integration.