Consider a problem below...
The solution offered to this particular question (1)a)) simply state the change of variable ksi to by to yield the result, I'm failing miserably to see how.
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First of all, write formula from $\textbf{[1]}$ for $x=yb$, that means:
$$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; d\xi =\pi e^{-b|y|}$$
Next change of variables in integral:
$$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; d\xi$$
Putting $\xi=\frac{\xi}{b}$:
$$\int_{-\infty}^{\infty}\frac{e^{i \xi yb}}{1+\xi^2}\; d\xi=\int_{-\infty}^{\infty}\frac{1}{b}\frac{e^{i \xi y}}{1+\frac{\xi^2}{b^2}}\; d\xi$$
Next (simple multiplying by $\frac{b}{b}=1$):
$$\int_{-\infty}^{\infty}\frac{1}{b}\frac{e^{i \xi y}}{1+\frac{\xi^2}{b^2}}\; d\xi=\int_{-\infty}^{\infty}b\frac{e^{i \xi y}}{b^2+\xi^2}\; d\xi$$
agha
- 10,038
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I suggest taking $s = \frac{\xi}{b}$ so the denominator looks as the denominator of the integral you know how to solve. Then, think using the condition "for any $x \in \mathbb{R} $".
