Laplace transform III

Laplace transform of the Sinhc and Sinc functions

We show the proof of the Laplace transforms of these slight variations of the Sinc and the Sinhc function from two problems posted by @BossGercek here and here. Both can be obtained from each other if we allow the parameters $\displaystyle b,c$ to have complex values. \[ \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = \ln\left(\sqrt{\frac{a+b}{a-b}}\right) \] \[ \int_{0}^{\infty} \frac{e^{-ax}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{a+ic}{a-ic}}\right)\] Taking care where the inverse hyperbolic and trigonometric functions are defined, if we put $b=c=1$ we get the classic values for the Laplace transform of the Sinc and Sinhc functions: \[ \int_{0}^{\infty} \frac{e^{-ax}\sinh x}{x} dx = \operatorname{arccoth} (a) \] \[ \int_{0}^{\infty} \frac{e^{-ax}\sin x}{x} dx = \frac{\pi}{2} - \arctan(a) = \operatorname{arccot}(a) \] If we put $a=1$ and maintain $b,c$ as free parameters we have: \[ \int_{0}^{\infty} \frac{e^{-x}\sinh bx}{x} dx = \operatorname{arctanh}(b) \] \[ \int_{0}^{\infty} \frac{e^{-x}\sin cx}{x} dx = \arctan(c) \] From the above we have the particular value: \[ \int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = \frac{\pi}{4} \]

Proof

Differtiating under the integral sign:
\[ \frac{d}{da} \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = -\int_{0}^{ \infty} e^{-ax}\sinh bx dx = - \left\{\mathscr{L} \sinh bx \right\}\left(a\right) \]
Hence
\begin{align*} \left\{\mathscr{L} \sinh bx \right\}\left(a\right) =& \int_{0}^{\infty } e^{-ax} \left(\frac{e^{bx}-e^{-bx}}{2}\right)dx \\ = & \frac{1}{2} \int_{0}^{\infty } e^{-x(a-b)} dx - \frac{1}{2} \int_{0}^{\infty } e^{-x(a+b)} dx \\ =& \frac{1}{2}\left(\frac{1}{a-b} - \frac{1}{a+b}\right)\\ \end{align*}
therefore
\[ \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = -\int \frac{1}{2}\left(\frac{1}{a-b} - \frac{1}{a+b}\right)da + C = \ln\left(\sqrt{\frac{a+b}{a-b}}\right) +C \]
Taking the limit as $b \to 0 \Longrightarrow \sin bx \to 0 \Longrightarrow C = 0 $
\[\boxed{ \int_{0}^{\infty} \frac{e^{-ax}\sinh bx}{x} dx = \ln\left(\sqrt{\frac{a+b}{a-b}}\right)} \]
For the Lapalce transform of the Sinc function:

Recall that
\[\sin x = -i \sinh ix \]
Therefore if we allow $\displaystyle b = ic$ :
\[ \int_{0}^{\infty} \frac{e^{-ax}\sinh ic x}{x} dx = \ln \left(\sqrt{\frac{a+ic}{a-ic}}\right) \Longrightarrow -i\int_{0}^{\infty} \frac{e^{-ax}\sinh ic x}{x} dx = -i\ln\left(\sqrt{\frac{a+ic}{a-ic}}\right)\]
Hence \[ \boxed{\int_{0}^{\infty} \frac{e^{-ax}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{a+ic}{a-ic}}\right)}\]

Now we will show a particular value:

Recall that in the complex plane $ \displaystyle \arctan(z) = -\frac{i}{2}\ln\left(\frac{1+iz}{1-iz}\right) $ Hence if $\displaystyle a=1$ \[ \int_{0}^{\infty} \frac{e^{-x}\sin cx}{x} dx = -i\ln \left(\sqrt{\frac{1+ic}{1-ic}}\right) = \arctan(c) \] Moreover, if $c=1$ \[ \int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = -i\ln \left(\sqrt{\frac{1+ic}{1-ic}}\right) = \arctan(1) = \frac{\pi}{4} \] Therefore \[ \boxed{ \int_{0}^{\infty} \frac{e^{-x}\sin x}{x} dx = \frac{\pi}{4} }\]