Evaluación de integrales utilizando la transformada inversa de Laplace. Laplace.

Problemas resueltos

Problema 1. Calcular $\displaystyle \int_{0}^{t}{J_0 (t) \ J_0 (t-u) \ du}$ .

Solución. Sea $\displaystyle g(t) = \int_{0}^{t}{J_0 (u) \ J_0 (t-u) \ du}$ . Luego, aplicando el teorema de convolución

$\displaystyle \mathcal{L}[g(t)] = \mathcal{L} [J_0 (t)] \ \mathcal{L} [J_0 (t)]$

$\displaystyle \mathcal{L}[g(t)] = \left(\frac{1}{\sqrt{s^2+1}} \right) \left(\frac{1}{\sqrt{s^2+1}} \right)$

$\displaystyle \mathcal{L}[g(t)] = \left(\frac{1}{\sqrt{s^2+1}} \right)^2$

$\displaystyle \mathcal{L}[g(t)] = \frac{1}{s^2+1}$

Despejando $g(t)$

$\displaystyle g(t) = \mathcal{L}^{-1} \left[\frac{1}{s^2+1} \right]$

$\displaystyle g(t) = \sin{t}$

$\displaystyle \int_{0}^{t}{J_0 (t) \ J_0 (t-u) \ du} = \sin{t}$

Por lo tanto,

$\displaystyle \therefore \int_{0}^{t}{J_0 (t) \ J_0 (t-u) \ du} = \sin{t}$

Problema 2. Hallar $\displaystyle \int_0^{\infty}{\cos{x^2} \ dx}$ .

Solución. Sea $\displaystyle f(t) = \int_0^{\infty}{\cos{tx^2} \ dx}$ . Después, aplicando la transformada de Laplace, se tiene que

$\displaystyle f(t) = \int_0^{\infty}{\cos{tx^2} \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \mathcal{L} \left[\int_0^{\infty}{\cos{tx^2} \ dx} \right]$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{e^{-st} \ dt \ \int_0^{\infty}{\cos{tx^2} \ dx}}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{dx \ \int_0^{\infty}{e^{-st} \ \cos{tx^2} \ dt}}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\mathcal{L} [\cos{tx^2}] \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\frac{s}{s^2 + (x^2)^2} \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\frac{s}{s^2 + x^4} \ dx}$

Ahora, sea $x^2= s \tan{\theta}$ , $\displaystyle x = \sqrt{s} \sqrt{\tan{\theta}}$ y $\displaystyle dx= \frac{\sqrt{s} \sec^2{\theta}}{2 \sqrt{\tan{\theta}}} \ d\theta$ . Entonces,

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\frac{s}{s^2 + x^4} \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\pi/2}{\frac{s}{s^2 + s^2 \tan^2{\theta}} \ \frac{\sqrt{s} \sec^2{\theta}}{2 \sqrt{\tan{\theta}}} \ d\theta}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\pi/2}{\frac{1}{1 + \tan^2{\theta}} \ \frac{\sec^2{\theta}}{2\sqrt{s} \sqrt{\tan{\theta}}} \ d\theta}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \int_0^{\pi/2}{\frac{1}{\sqrt{\tan{\theta}}} \ d\theta}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \int_0^{\pi/2}{\tan^{-1/2}{\theta} \ d\theta}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \int_0^{\pi/2}{\sin^{-1/2}{\theta} \cos^{1/2}{\theta} \ d\theta}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \cdot \frac{\Gamma \left(\frac{1}{4} \right) \Gamma \left(\frac{3}{4} \right)}{2 \Gamma \left(\frac{1}{4} + \frac{3}{4} \right)}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \cdot \frac{\Gamma \left(\frac{1}{4} \right) \Gamma \left(\frac{3}{4} \right)}{2 \Gamma \left(1 \right)}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{2 \sqrt{s}} \cdot \frac{\frac{\pi}{\sin{\frac{\pi}{4}}}}{2 \left(1 \right)}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{4 \sqrt{s}} \cdot \frac{\pi}{\sin{\frac{\pi}{4}}}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{4 \sqrt{s}} \cdot \frac{\pi}{\frac{1}{\sqrt{2}}}$

$\displaystyle \mathcal{L} [f(t)] = \frac{\pi \sqrt{2}}{4 \sqrt{s}}$

Despejando $f(t)$

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{\pi \sqrt{2}}{4 \sqrt{s}} \right]$

$\displaystyle f(t) = \frac{\pi \sqrt{2}}{4} \mathcal{L}^{-1} \left[\frac{1}{\sqrt{s}} \right]$

$\displaystyle f(t) = \frac{\pi \sqrt{2}}{4} \left(\frac{t^{-1/2}}{\sqrt{\pi}} \right)$

$\displaystyle f(t) = \frac{\sqrt{2 \pi}}{4} t^{-1/2}$

$\displaystyle \int_0^{\infty}{\cos{tx^2} \ dx} = \frac{\sqrt{2 \pi}}{4} t^{-1/2}$

Si $t=1$ , el resultado final es

$\displaystyle \int_0^{\infty}{\cos{(1) x^2} \ dx} = \frac{\sqrt{2 \pi}}{4} {(1)}^{-1/2}$

$\displaystyle \int_0^{\infty}{\cos{x^2} \ dx} = \frac{\sqrt{2 \pi}}{4}$

$\displaystyle \therefore \int_0^{\infty}{\cos{x^2} \ dx} = \frac{1}{2} \sqrt{\frac{\pi}{2}}$

Problema 3. Hallar $\displaystyle \int_{0}^{\infty}{e^{-x^2} \ dx}$ .

Solución. Sea $\displaystyle f(t) = \int_{0}^{\infty}{e^{-tx^2} \ dx}$ . Aplicando la transformada de Laplace, resulta

$\displaystyle \mathcal{L} [f(t)] = \mathcal{L} \left[\int_{0}^{\infty}{e^{-tx^2} \ dx} \right]$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{e^{-st} \ dt \ \int_{0}^{\infty}{e^{-tx^2} \ dx}}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{dx \ \int_{0}^{\infty}{e^{-st} \ e^{-tx^2} \ dt}}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\mathcal{L} [e^{-tx^2}] \ dx} = \int_0^{\infty}{\mathcal{L} [e^{-x^2 t}] \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \int_0^{\infty}{\frac{1}{s + x^2} \ dx}$

$\displaystyle \mathcal{L} [f(t)] = \left[\frac{1}{\sqrt{s}} \arctan{\left(\frac{x}{\sqrt{s}} \right)} \right]_0^{\infty}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{\sqrt{s}} \arctan{\left(\frac{\infty}{\sqrt{s}} \right)} - \arctan{\left(\frac{0}{\sqrt{s}} \right)}$

$\displaystyle \mathcal{L} [f(t)] = \frac{1}{\sqrt{s}} \left(\frac{\pi}{2} \right)$

$\displaystyle \mathcal{L} [f(t)] = \frac{\pi}{2 \sqrt{s}}$

Despejando $f(t)$ , se tiene que

$\displaystyle f(t) = \mathcal{L}^{-1} \left[\frac{\pi}{2 \sqrt{s}} \right]$

$\displaystyle f(t) = \frac{\pi}{2} \mathcal{L}^{-1} \left[\frac{1}{\sqrt{s}} \right]$

$\displaystyle f(t) = \frac{\pi}{2} \left(\frac{t^{-1/2}}{\sqrt{\pi}} \right)$

$\displaystyle f(t) = \frac{\sqrt{\pi}}{2} t^{-1/2}$

$\displaystyle \int_{0}^{\infty}{e^{-tx^2} \ dx} = \frac{\sqrt{\pi}}{2} t^{-1/2}$

Cuando $t=1$ , se tiene el resultado final

$\displaystyle \int_{0}^{\infty}{e^{-(1)x^2} \ dx} = \frac{\sqrt{\pi}}{2} {(1)}^{-1/2}$

$\displaystyle \therefore \int_{0}^{\infty}{e^{-x^2} \ dx} = \frac{\sqrt{\pi}}{2}$

Evaluación de integrales utilizando la transformada inversa de Laplace. Laplace.

Publicado por Cesar Reyes

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Evaluación de integrales utilizando la transformada inversa de Laplace. Laplace.

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Publicado por Cesar Reyes

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