How to rewrite $\frac{1}{-a+ib}$ with partial fractions?

Question

I faced some difficulties while trying to calculate $\int e^{-ax}$cos(bx)dx. I rewrote this as the integral of $e^{(-a + ib)x}$ calculating $\int e^{(-a + ib)x}dx$ (the answer is then the real part of the latter integral). I found $$\int e^{(-a + ib)x}dx = \frac{1}{-a+ib}e^{(-a + ib)x}.$$ My problem is that I don't know how to rewrite $\frac{1}{-a+ib}$, because when I try to use the basic rules for partial fractions, I get stuck. Thanks!

Answer 1

write $$\frac{1}{-a+bi}\cdot \frac{-a-bi}{-a-bi}$$

Answer 2

Using $$\int e^{(-a + ib)x}dx = \frac{1}{-a+ib}e^{(-a + ib)x}$$ and $e^{(-a + i b) x} = e^{-a x} \, (\cos(b x) + i \sin(b x))$ then \begin{align} \int e^{(-a + ib)x}dx &= \frac{1}{-a+ib}e^{(-a + ib)x} \\ &= \frac{1}{-a + i b} \cdot \frac{-a - i b}{-a - i b} \, e^{(-a + ib) x} \\ &= \frac{-a - i b}{a^{2} + b^{2}} \, e^{(-a + ib) x} \\ &= \frac{e^{-a x}}{a^{2} + b^{2}} \, [ (a \cos(bx) - b \sin(b x)) + i \, (b \cos(b x) + a \sin( b x))] \end{align} for which \begin{align} \int e^{-a x} \, \cos(b x) \, dx &= \frac{e^{-a x}}{a^{2} + b^{2}} \, (a \cos(bx) - b \sin(b x)) \\ \int e^{- a x} \, \sin(b x) \, dx &= \frac{e^{-a x}}{a^{2} + b^{2}} \, (b \cos(b x) + a \sin( b x)) \end{align}

Suppose the limits of the integral are $(0, \infty)$ then these become \begin{align} \int_{0}^{\infty} e^{-a x} \, \cos(b x) \, dx &= \frac{a}{a^{2} + b^{2}} \\ \int_{0}^{\infty} e^{- a x} \, \sin(b x) \, dx &= \frac{b}{a^{2} + b^{2}} \end{align}

Answer 3

First off, I think that the problem that you are having is with simplifying $$ \frac{1}{-a+ib}, $$ which is not a partial fractions problem at all. Instead, multiply upstairs and downstairs by the conjugate $-a-ib$, then simplify. That will finish the problem for you.

That said, starting from the initial integral, I get the following (without having to track real and imaginary parts, at the cost of doing a little extra computation): \begin{align} \int \mathrm{e}^{-ax} \cos(bx) \,\mathrm{d}x &= \frac{1}{2} \int \mathrm{e}^{-ax} \left( \mathrm{e}^{ibx} + \mathrm{e}^{-ibx}\right) \,\mathrm{d}x \\ &= \frac{1}{2} \int \mathrm{e}^{(-a+ib)x} + \mathrm{e}^{(-a-ib)x}\, \mathrm{d}x \\ &= \frac{1}{2} \left( \frac{1}{-a+ib}\mathrm{e}^{(-a+ib)x} + \frac{1}{-a-ib}\mathrm{e}^{(-a-ib)x}\right) \\ &= \frac{\mathrm{e}^{-ax}}{2} \left( \frac{1}{-a+ib}\cdot \frac{-a - ib}{-a - ib}\mathrm{e}^{ibx} - \frac{1}{a+ib}\frac{a - ib}{a - ib}\mathrm{e}^{-ibx}\right) \\ &= \frac{\mathrm{e}^{-ax}}{2(a^2+b^2)} \left( \left(-a-ib \right)\mathrm{e}^{ibx} - \left(a-ib \right)\mathrm{e}^{-ibx}\right) \\ &= \frac{\mathrm{e}^{-ax}}{2(a^2+b^2)} \left( -a\left( \mathrm{e}^{ibx} + \mathrm{e}^{ibx} \right) - ib\left( \mathrm{e}^{ibx} - \mathrm{e}^{ibx} \right)\right) \\ &= \frac{\mathrm{e}^{-ax}}{2(a^2+b^2)} \left( -2a\left( \frac{\mathrm{e}^{ibx} + \mathrm{e}^{ibx}}{2} \right) - (2i)ib\left( \frac{\mathrm{e}^{ibx} - \mathrm{e}^{ibx}}{2i} \right)\right) \\ &= \frac{\mathrm{e}^{-ax}}{2(a^2+b^2)} \left( -2a\cos(bx) + 2b \sin(bx) \right) \\ &= \frac{\mathrm{e}^{-ax}(b\sin(bx) - a\cos(bx))}{(a^2+b^2)}. \end{align}

That being said, I think that it might be more straight-forward to tackle this integral via the usual integration by parts trickery: \begin{align} I &= \int \mathrm{e}^{-ax} \cos(bx)\,\mathrm{d}x \\ &= \frac{\mathrm{e}^{-ax}\sin(bx)}{b} + \frac{a}{b} \int \mathrm{e}^{-ax}\sin(bx)\,\mathrm{d}x \\ &= \frac{\mathrm{e}^{-ax}\sin(bx)}{b} - \frac{a \mathrm{e}^{-ax} \cos(bx)}{b^2} - \frac{a^2}{b^2} \int \mathrm{e}^{-ax}\cos(bx)\,\mathrm{d}x \\ &= \frac{\mathrm{e}^{-ax}\sin(bx)}{b} - \frac{a \mathrm{e}^{-ax} \cos(bx)}{b^2} - \frac{a^2}{b^2}I. \end{align} Solving for $I$, we obtain \begin{align} \left(1+\frac{a^2}{b^2}\right)I = \frac{\mathrm{e}^{-ax}\sin(bx)}{b} - \frac{a \mathrm{e}^{-ax} \cos(bx)}{b^2} &\implies \left( \frac{b^2 + a^2}{b^2} \right) I = \frac{\mathrm{e}^{-ax}}{b^2} \left( b \sin(bx) - a \cos(bx) \right) \\ &\implies I = \frac{\mathrm{e}^{-ax}}{a^2 + b^2} \left( b \sin(bx) - a \cos(bx) \right), \end{align} which is the same as the above, but requires no complexification.