Question

(a) Show that $\delta[(t-a_1)(t-a_2)] = \frac{1}{|a_1 - a_2|}[\delta(t-a_1) + \delta(t-a_2)]$ (b) Evaluate $\int_{-\infty}^{+\infty} (2t^3 - 3t) \exp(-\frac{t^2}{2}) \frac{d}{dt} \{\delta(t^2 - 5t + 6)} dt.$

          (a) Show that
$\delta[(t-a_1)(t-a_2)] = \frac{1}{|a_1 - a_2|}[\delta(t-a_1) + \delta(t-a_2)]$
(b) Evaluate
$\int_{-\infty}^{+\infty} (2t^3 - 3t) \exp(-\frac{t^2}{2}) \frac{d}{dt} \{\delta(t^2 - 5t + 6)} dt.$

$(a) Show that δ[(t-a1)(t-a2)] = (1)/(|a1 - a2|)[δ(t-a1) + δ(t-a2)] (b) Evaluate ∫-∞^+∞ (2t^3 - 3t) (-(t^2)/(2)) (d)/(dt){δ(t^2 - 5t + 6) dt.$

Added by Samuel M.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Samriddhi Singh

03/03/2022

Step 1

For a smooth function g(t) with simple zeros t_i (g(t_i)=0 and g'(t_i)≠0), the distributional identity holds: ∫ φ(t) δ(g(t)) dt = ∑_i φ(t_i)/|g'(t_i)| for every test function φ, so δ(g(t)) = ∑_i δ(t - t_i)/|g'(t_i)|. Show more…

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