Integration of complex exponential $\int_{\alpha}^{\beta}e^{cx}dx=\frac{1}{c}(e^{c\beta}-e^{c\alpha})$

Question

If $c=a+bi$ is a complex constant, show that $$\int_{\alpha}^{\beta}e^{cx}dx=\frac{1}{c}(e^{c\beta}-e^{c\alpha})$$ by writing out the real and imaginary parts of both sides.

I presume this isn't as simple as taking a regular antiderative of $e^{cx}$ and there is some nuance here with the complex numbers. I have never worked with complex numbers outside of basic properties, so I am not really sure how to work this problem. Does the problem want me to break up the integral of $e^{ax}$ and $e^{bix}$ then perhaps show the product of these integrals equals the above?

Answer 1

The problem asks to write out explicitly the real and imaginary parts of the two sides, and then show that they are equal respectivley.

First, write out the RHS

$$RHS=\frac{e^{(a+ib)\beta}-e^{(a+ib)\alpha}}{a+bi} = \frac{(a-bi)[e^{(a+ib)\beta}-e^{(a+ib)\alpha}]}{a^2+b^2}$$ $$= \frac{(a-bi)[(e^{a\beta}\cos b\beta -e^{a\alpha}\cos b\alpha) + i(e^{a\beta}\sin b\beta - e^{a\alpha}\sin b\alpha)]}{a^2+b^2}$$ $$= RHS_{Re}+iRHS_{Im}$$

where the real and imaginary parts are respectively,

$$RHS_{Re}= \frac{e^{a\beta}(a\cos b\beta +b\sin b\beta ) - e^{a\alpha}(a\cos b\alpha+b\sin b\alpha)}{a^2+b^2}\tag 1$$ $$RHS_{Im}= \frac{e^{a\beta}(a\sin b\beta-b\cos b\beta ) - e^{a\alpha}(a\sin b\alpha-b\cos b\alpha)}{a^2+b^2}\tag 2$$

Then, write out the LHS,

$$LHS=\int_{\alpha}^{\beta}e^{(a+bi)x}dx= \int_{\alpha}^{\beta} e^{ax} (\cos bx+i\sin bx)dx$$

Integrate the real part with integration by parts,

$$LHS_{Re}= \int_{\alpha}^{\beta} e^{ax} \cos bx \> dx =\frac1a \int_{\alpha}^{\beta}d( e^{ax}) \cos bx \> dx$$

$$=\frac1a (e^{a\beta} \cos b\beta - e^{a\alpha} \cos b\alpha) +\frac b{a^2} \int_{\alpha}^{\beta}d( e^{ax}) \sin bx \> dx$$

$$=\frac1a (e^{a\beta} \cos b\beta - e^{a\alpha} \cos b\alpha) +\frac b{a^2} (e^{a\beta} \sin b\beta - e^{a\alpha} \sin b\alpha) -\frac {b^2}{a^2} LHS_{Re}$$

Rearrange to have

$$LHS_{Re}= \frac{e^{a\beta}(a\cos b\beta +b\sin b\beta ) - e^{a\alpha}(a\cos b\alpha+b\sin b\alpha)}{a^2+b^2}\tag 3$$

Similarly, integrate the imaginary part,

$$LHS_{Im}= \int_{\alpha}^{\beta} e^{ax} \sin bx \> dx =\frac1a \int_{\alpha}^{\beta}d( e^{ax}) \sin bx \> dx$$

$$=\frac1a (e^{a\beta} \sin b\beta - e^{a\alpha} \sin b\alpha) -\frac b{a^2} \int_{\alpha}^{\beta}d( e^{ax}) \cos bx \> dx$$

$$=\frac1a (e^{a\beta} \sin b\beta - e^{a\alpha} \sin b\alpha) -\frac b{a^2} (e^{a\beta} \cos b\beta - e^{a\alpha} \cos b\alpha) -\frac {b^2}{a^2} LHS_{Re}$$

or,

$$LHS_{Im}= \frac{e^{a\beta}(a\sin b\beta-b\cos b\beta ) - e^{a\alpha}(a\sin b\alpha-b\cos b\alpha)}{a^2+b^2}\tag 4$$

Thus, as seen from (1), (3) and (2), (4), the real and imaginary parts of the two sides are equal respectively

$$LHS_{Re}= RHS_{Re},\>\>\>\>\> LHS_{Im }= RHS_{Im}$$

Answer 2

This is what they are looking for. Use the fact that $e^{cx}=e^{ax+bxi}=e^{ax}(\cos(bx)+i\sin(bx)).$ Then the integral becomes $$ \int_\alpha^\beta e^{cx}\,\mathrm{d}x=\int_\alpha^\beta e^{ax}\cos(bx)\,\mathrm{d}x+i\int_\alpha^\beta e^{ax}\sin(bx)\,\mathrm{d}x. $$ From here, use the well-known "bounce back" technique (integration by parts twice) to evaluate both integrals. Then recombine and you should get $\frac{1}{c}(e^{c\beta}-e^{c\alpha})$ after simplifying and suppressing the notation.