This integral arose in a problem where I needed to apply Fourier transform of the impulse response. I have $$\int\limits_{-\infty}^{\infty}u(t)e^{-(b+j\omega)t} \ dt, \quad u(t)=1 \ \text{if} \ t>0 \ \text{and}\ 0 \ \text{otherwise.}\tag1$$
Using integration by parts I then get
$$u(t)\int\limits_{-\infty}^{\infty}e^{-(b+j\omega)t} \ dt - \int\limits_{-\infty}^{\infty}\delta(t)\underbrace{\left[\int\limits_{-\infty}^{\infty}e^{-(b+j\omega)t} \ dt\right]}_{=f(t)} \ dt. \tag2$$
Where $u'(t)=\delta(t).$ Now, in the first integral, $u(t)=0$ for $t<0$ so the contributions are only from $t>0$ and we can adjust the limits accordingly. So $(2)$ becomes
$$\int\limits_{0}^{\infty}e^{-(b+j\omega)t} \ dt-\int\limits_{-\infty}^{\infty}\delta(t)f(t) \ dt = \left[-\frac{e^{-(b+j\omega)t}}{b+j\omega}\right]^{\infty}_{0}-f(0) = \frac{1}{b+j\omega}-f(0).\tag3$$
What is $f(0)$? Plugging in $t=0$ I just get $f(0)=\infty$. According to my book the correct answer is just $1/(b+j\omega).$
Note that to evaluate the second integral in $(2)$ I wanted to use the well known fact that $$\int\limits_{-\infty}^{\infty}\delta(t)f(t) \ dt = f(0).$$
I think you’ve mistakenly calculated $e^0$ as 0 instead of 1 with regards to $f(0)$.