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Don’t know how to answer this question.

please give the answer if anyone knows..

I am using this equation

\begin{align}

\sqrt{\frac{2}{\pi}}\int^\infty_0 f(x) \cos (\omega x) \,dx

\end{align}

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Let $F(\omega)$ be represented by the integral

$$F(\omega)=\sqrt{\frac 2\pi}\int_0^\infty \frac{\cos(\omega x)}{1+x^2}\,dx\tag1$$

Next, given that $\int_0^\infty \frac{x\sin(\omega x)}{1+x^2}\,dx$ converges uniformly for $\omega\ge \omega_0>0$ then

$$\begin{align}

F'(\omega)&=-\sqrt{\frac 2\pi}\int_0^\infty \frac{x\sin(\omega x)}{1+x^2}\,dx\\\\

&=-\sqrt{\frac 2\pi}\int_0^\infty \frac{(1+x^2-1)\sin(\omega x)}{x(1+x^2)}\,dx\\\\

&=-\sqrt{\frac 2\pi}\left(\frac\pi 2 -\int_0^\infty \frac{\sin(\omega x)}{x(1+x^2)}\right)\,dx\tag 2

\end{align}$$

And owing to uniform convergence again, we assert that

$$\begin{align}

F”(\omega)&=\sqrt{\frac 2\pi}\int_0^\infty \frac{\cos(\omega x)}{1+x^2}\,dx\\\\

&=F(\omega)\tag3

\end{align}$$

Solving the ODE in $(3)$ yields

$$F(\omega)=Ae^{\omega}+Be^{-\omega} \tag 4$$

Now, letting $\omega \to 0$ in $(1)$ and $(2)$ reveals that $F(0)=\sqrt{\frac{\pi}{2} }$ and $F'(0)=-\sqrt{\frac{\pi}{2}}$.

Applying these initial conditions to $(4)$ yields

$$F(\omega)=\sqrt{\frac{\pi}{2}}e^{-|\omega|}$$

And we are done!

Hint: Observe $f(x) = (1+x^2)^{-1}$ is even, which means

\begin{align}

\int^\infty_0 f(x)\cos \omega x\ dx =&\ \frac{1}{2}\int^\infty_{-\infty} f(x)\cos\omega x\ dx

=\ \frac{1}{2}\int^\infty_{-\infty} f(x)\frac{e^{i\omega x}+e^{-i\omega x}}{2}\ dx\\

=&\ \frac{1}{4}\int^\infty_{-\infty} f(x)e^{i\omega x}\ dx+\frac{1}{4}\int^\infty_{-\infty}f(x)e^{-i\omega x}\ dx= \frac{1}{2}\int^\infty_{-\infty} f(x) e^{-i\omega x}\ dx.

\end{align}

A hint: You have seen Dr. MV’s answer. If you are not convinced compute the Fourier transform of $F(\omega):=\sqrt{\pi/2}\>e^{-|\omega|}$. You should obtain the given $f(x)$.

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