Complex exponent integral - prove $\int_a^b{e^{\lambda x} \text{dx}}=\frac{1}{\lambda}\left(e^{\lambda b}-e^{\lambda a}\right) $

Question

How to prove the exponent integration rule: $$\int_a^b{e^{\lambda x} \text{dx}}=\frac{1}{\lambda}\left(e^{\lambda b}-e^{\lambda a}\right) $$ In the complex version of it - that is, when $\lambda \neq 0$ and $\lambda \in \Bbb C$, under the complex function derivative definition: $f(x)=u(x)+iv(x) \rightarrow f'(x)=u'(x)+iv'(x)$?

Answer 1

You should show that for $\lambda \in \mathbb{C}$ and $\lambda \neq 0$: $$(e^{\lambda x})'=\frac{e^{\lambda x}}{\lambda}$$

But if you write $\lambda=a+ib$ for $a,b \in \mathbb{R}$ it's clear, because:$$(e^{\lambda x})'=(e^{(a+ib)x})'=(e^{ax}e^{ibx})'=(e^{ax}(\cos bx+ i \sin x))'=(e^{ax}\cos bx+ie^{ax}\sin bx)'$$

Now calculate derivatives of $e^{ax}\cos bx$ and $e^{ax}\sin bx$ to use your formula.