So, one of my homework assignments is to take the Laplace transform of a function such that $f(t)=\cosh{bt}$. I figured it would be equivalent to:
$$\int^\infty_0{e^{-st}\frac{e^{bt}+e^{-bt}}{2}}\mathrm{dt}=\frac{1}{2}(\int^\infty_0{e^{t(b-s)}}\mathrm{dt}+\int^\infty_0{e^{-t(b+s)}}\mathrm{dt})=\frac{1}{2}\left(\frac{e^{t(b-s)}}{2(b-s)}+\frac{e^{-t(b+s)}}{2(b+s)}\right)\Bigg]^{\infty}_0$$
However the first term of the result is equal to $\infty$ because we have the exponential function over a constant. Could someone be so kind as to point out my mistake?
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$\begingroup$You made a few mistakes when you integrated, like the appearance of an extra $\frac{1}{2}$ coefficient, and forgetting to include a negative when you integrate $\int^\infty_0{e^{-t(b+s)}}\mathrm{dt}.$ After integration you should get $$\frac{e^{t(b-s)}}{2(b-s)}-\frac{e^{-t(b+s)}}{2(b+s)} \Bigg|_0^\infty$$ The next move (I believe) is that you assume $\text{Re}(s)>b$. In that way, $e^{t(b-s)} $ will converge to $0$ as $t \to \infty$. So, $$\frac{e^{t(b-s)}}{2(b-s)}-\frac{e^{-t(b+s)}}{2(b+s)} \Bigg|_0^\infty = -\frac{1}{2(b-s)}+\frac{1}{2(b+s)} \\ = \frac{s}{s^2-b^2}$$
$\endgroup$ 2 $\begingroup$The limit: $$ \lim_{t \to +\infty} \frac {e^{t(b-s)}}{2(b-s)}$$ exists if $\operatorname{Re}(s) > b$.
This might be more obvious if you write it as:
$$ \lim_{t \to +\infty} \frac {e^{-t(s-b)}}{2(b-s)}$$
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