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$$ \int_{0}^{\infty}\frac{t^{1/2}}{1+t^2} dt$$

Must use the following to evaluate the integral above:

$$ \int_{0}^{\infty}\exp{(-y^2)} dy = \frac{\sqrt{\pi}}{2}$$

HINTS

  • Use of Parseval Theorem
  • $\mathcal{F_{s}[t^{-1/2}]} = \omega ^{-1/2}$
  • Fourier Sine/Cosine table

FUNDAMENTAL INFORMATION

$$\mathcal{F_{c}[f(t)]} = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} f(t) \cos{\omega t}dt$$ $$\mathcal{F_{s}[f(t)]} = \sqrt{\frac{2}{\pi}}\int_{0}^{\infty} f(t) \sin{\omega t}dt$$

FINAL EVALUATION

$$ \int_{0}^{\infty}\frac{t^{1/2}}{1+t^2} dt = \frac{\pi}{\sqrt{2}}$$

Want the answer through the FST/FCT.

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You have all the ingredients you need. The Fourier transform of $\frac{1}{1+t^2}$ gives you an exponential function (by the relation between the Cauchy distribution and the Laplace distribution), the Fourier transform of $\sqrt{t}$ gives you a $\frac{1}{\sqrt{\omega}}$, apply the change of variable $\omega=x^2$ and you are done.

You may also tackle the problem through the Laplace transform. Namely: $$ \int_{0}^{+\infty}\mathcal{L}\left(\frac{1}{\sqrt{t}}\right)(s)\cdot\mathcal{L}^{-1}\left(\frac{t}{1+t^2}\right)(s)\,ds = \sqrt{\pi}\int_{0}^{+\infty}\frac{\cos(s)}{\sqrt{s}}\,ds =\sqrt{\pi}\underbrace{\int_{-\infty}^{+\infty}\cos(u^2)\,du}_{\text{Fresnel integral}.}$$

You may also invoke Glasser's master theorem: $$ \int_{0}^{+\infty}\frac{\sqrt{t}}{1+t^2}\,dt = \int_{-\infty}^{+\infty}\frac{du}{\left(u-\frac{1}{u}\right)^2+2}=\int_{-\infty}^{+\infty}\frac{dv}{v^2+2}=\color{blue}{\frac{\pi}{\sqrt{2}}}. $$

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