“Proof” that the Laplace Transform is Nothing

-March 27, 2015

Can you find the error(s), if any, in the following mathematical derivation that applies the Laplace transform? Begin by expressing the function to be transformed, g(t), in the form of the square of (t), or g(t) = 2(t). Then by the definition of the Laplace transform,

Apply the integration by parts formula for definite integrals:

where u and v are both functions of t, and ti and tf are the integration limits of t. Choose

and

Combining these in the formula,

where the integration limits that applied to t are converted to f. To convert them, apply the condition on (t) that

Thus,

The integration is with respect to f, not t, and consequently the limits of integration have been changed to (0) and (∞), as indicated. When the integral is taken and the resulting expression is simplified,

Therefore, the Laplace transform of all functions of the form g(t) are G(s) = 0.

Nobody who has worked with Laplace transforms would accept this result as valid. Where is the error?

Hint: one error is in the line that reads

To find v = v(t), the indefinite integral of the integrand f(t)∙est must be taken, then the limits of integration applied to it. The error is in equating expressions that are not necessarily equal:

If v were equal to F(s), then v(t) is constant with respect to F(s) and dv/dt = 0, which is consistent with the faulty conclusion.

Erik Margan, sensor-electronics designer of the ATLAS experiment of the Large Hadron Collider (LHC) at CERN found additional errors. Can you identify them?

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