What is the value of $\displaystyle \int_{-\infty}^\infty f(x)f''(x)\,dx$ where $f(x)=\dfrac{\alpha}{(x^2+\beta^2)},\quad \alpha,\beta \in \mathbb{R}$? I have tried to use the Fourier transform of $f(x)$ which is an exponential function, but stuck somewhere?
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Am I right : $f'(x) = \frac{-2x\beta}{(x^{2}+\beta^{2})^{2}}$? Or it should derivative with respect to $\beta$? – openspace Apr 04 '18 at 04:50
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yes. derivative with respect to x and not \beta. – Purushothaman Apr 04 '18 at 08:32
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Integrating by parts we have $\int_{-\infty} ^{\infty} f(x)f''(x)dx=-\int_{-\infty} ^{\infty} (f'(x))^{2}dx $ $= -\int_{-\infty} ^{\infty}\frac {(2\alpha x)^{2}} {(x^{2}+\beta^{2})^{2}}$ dx. You can easily evaluate this bu the substitution $x=\beta \tan(\theta)$.
Kavi Rama Murthy
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