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Sunday, August 22, 2021

Laplace transform III

Sinc and Sinch functions

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 b,c to have complex values. 0eaxsinhbxxdx=ln(a+bab)
0eaxsincxxdx=iln(a+icaic)
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: 0eaxsinhxxdx=arccoth(a)
0eaxsinxxdx=π2arctan(a)=arccot(a)
If we put a=1 and maintain b,c as free parameters we have: 0exsinhbxxdx=arctanh(b)
0exsincxxdx=arctan(c)
From the above we have the particular value: 0exsinxxdx=π4


Proof

Differtiating under the integral sign:
dda0eaxsinhbxxdx=0eaxsinhbxdx={Lsinhbx}(a)

Hence
{Lsinhbx}(a)=0eax(ebxebx2)dx=120ex(ab)dx120ex(a+b)dx=12(1ab1a+b)

therefore
0eaxsinhbxxdx=12(1ab1a+b)da+C=ln(a+bab)+C

Taking the limit as b0sinbx0C=0
0eaxsinhbxxdx=ln(a+bab)

For the Lapalce transform of the Sinc function:

Recall that
sinx=isinhix

Therefore if we allow b=ic :
0eaxsinhicxxdx=ln(a+icaic)i0eaxsinhicxxdx=iln(a+icaic)

Hence 0eaxsincxxdx=iln(a+icaic)


Now we will show a particular value:

Recall that in the complex plane arctan(z)=i2ln(1+iz1iz) Hence if a=1 0exsincxxdx=iln(1+ic1ic)=arctan(c)
Moreover, if c=1 0exsinxxdx=iln(1+ic1ic)=arctan(1)=π4
Therefore 0exsinxxdx=π4

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