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Using laplace theorem find the answer

Using laplace theorem find the answer-example-1

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Here's the general approach.

For the first integral, assume
t>0. I'm also interpreting it to say


I_1=\displaystyle\int_0^\infty\sin(tx^2)\,\mathrm dx

(otherwise the integral simply diverges).

Writing
I_1=I_1(t), take the Laplace transform to get


\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty\left(\int_0^\infty\sin(tx^2)\,\mathrm dx\right)e^(-st)\,\mathrm dt

\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty\left(\underbrace{\int_0^\infty \sin(tx^2)e^(-st)}_{\mathcal L_s\{\sin(tx^2)\}}\,\mathrm dt\right)\,\mathrm dx

\mathcal L_s\{I_1(t)\}=\displaystyle\int_0^\infty(x^2)/(s^2+x^4)\,\mathrm dx

which is an elementary integral that can be computed by decomposing into partial fractions, or by employing a proper trigonometric substitution. I leave that calculation to you; you should end up with


\mathcal L_s\{I_1(t)\}=\frac\pi{2√(2s)}

and taking the inverse Laplace transform yields the answer,


I_1(t)=\mathcal L_t^(-1)\left\{\frac\pi{2√(2s)}\right\}=(\sqrt\pi)/(2√(2t))

Hopefully that should give you a decent idea of how this method works.

For the remaining integrals, you will need to introduce a parameter before proceeding. Here's what you can try:


I_2(t)=\displaystyle\int_0^\infty\frac{\sin(tx)}x\,\mathrm dx

I_3(t)=\displaystyle\int_0^\infty\frac{1-\cos(tx)}x\,\mathrm dx

I_4(t)=\displaystyle\int_0^\infty\frac{\sin^2(tx)}x\,\mathrm dx

For the fifth integral, you can make the observation that the given integrand already resembles the Laplace transform of
\frac{\sin t}t:


\underbrace{\mathcal L_s\left\{\frac{\sin t}t\right\}}_(Y(s))=\displaystyle\int_0^\infty e^(-st)\underbrace{\frac{\sin t}t}_(y(t))\,\mathrm dt\implies \int_0^\infty e^(-4t)\frac{\sin t}t\,\mathrm dt=Y(4)

Finally, for the last integral, you can use


I_6(t)=\displaystyle\int_0^\infty e^(-tx^2)\,\mathrm dx
User Jari Jokinen
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