The Laplace transform of the function \(\sqrt{\frac{t}{\pi}}\cos(8t)\) can be derived using the individual transforms of its components. The Laplace transform \(L\{\sqrt{\frac{t}{\pi}}\}\) is \(\frac{1}{s^{3/2}}\) and \(L\{\cos(8t)\}\) is \(\frac{s}{s^2 + 64}\). To find the Laplace transform of the product, one must compute the convolution of these two transforms, represented as \((f * g)(t)\). This involves evaluating the integral \(\frac{1}{\sqrt{\pi}} \int_0^t (t-v)^{1/2} \cos(8v) dv\).
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Saladsamurai said:Do you know what L\{\sqrt{{t}/{\pi}}\} is? Do you know what L\{\cos(8t)\} is?
jackmell said:I believe that requires evaluating a convoluted contour integral, which interestingly, equates to just it's residues but (rigorously) showing that seems pretty tough. I was just curious if the OP was prepared to do that type of analysis or if there is a more elementary way?
Econometricia said:Yes, I do know those transforms. I am now trying to express the integral as a gamma function.
jackmell said:I believe that requires evaluating a convoluted contour integral, which interestingly, equates to just it's residues but (rigorously) showing that seems pretty tough. I was just curious if the OP was prepared to do that type of analysis or if there is a more elementary way?
I am finding the Laplace transform of t^(1/2) in tables without use of the gamma function (there is a 'pi' factor in it which suggests it has already been evaluated for us). So it seems that iffzero said:The integral can be expressed in terms of a gamma function.
<br /> L\{\sqrt{{t}/{\pi}}\} = \frac{1}{s^(3/2)}<br />Saladsamurai said:Why don't you show what you have done so that we can better assist you?
I am finding the Laplace transform of t^(1/2) in tables without use of the gamma function (there is a 'pi' factor in it which suggests it has already been evaluated for us). So it seems that if
<br /> L\{\sqrt{{t}/{\pi}}\} = f(t)<br />
and
<br /> L\{\cos(8t)\} = g(t)<br />Then we need to find (f * g)(t); that is, find the "convolution" of f(t) and g(t).