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Question Number 132708 by frc2crc last updated on 16/Feb/21

$${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}\frac{x^2\cos (px+q)}{x^2+{(p+q)}^2}dx$$

Answered by Olaf last updated on 16/Feb/21

$$$$$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2\cos (px+q)}{x^2+{(p+q)}^2}dx$$$$\mathrm{\Omega }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2[\cos \left(px\right)\cos q-\sin \left(px\right)\sin q]}{x^2+{(p+q)}^2}dx$$$$\mathrm{\Omega }=\cos q{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{x^2\cos \left(px\right)}{x^2+{(p+q)}^2}dx$$$$...$$$$\mathrm{now}\mathrm{see}\mathrm{Q}.132090\mathrm{by}\mathrm{rs}4090$$

Answered by mathmax by abdo last updated on 16/Feb/21

$$\mathrm{\Phi }={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{x}}^2\cos (\mathrm{px}+\mathrm{q})}{{\mathrm{x}}^2+{(\mathrm{p}+\mathrm{q})}^2}\mathrm{dx}\mathrm{wetake}\mathrm{p}+\mathrm{q}>0\Rightarrow \mathrm{\Phi }{=}_{\mathrm{x}=(\mathrm{p}+\mathrm{q})\mathrm{t}}$$$$={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{(\mathrm{p}+\mathrm{q})}^2{\mathrm{t}}^2\cos (\mathrm{p}(\mathrm{p}+\mathrm{q})\mathrm{t}+\mathrm{q})}{{(\mathrm{p}+\mathrm{q})}^2({\mathrm{t}}^2+1)}(\mathrm{p}+\mathrm{q})\mathrm{dt}$$$$=(\mathrm{p}+\mathrm{q}){\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^2\cos (({\mathrm{p}}^2+\mathrm{pq})\mathrm{t}+\mathrm{q})}{{\mathrm{t}}^2+1}\mathrm{dt}=(\mathrm{p}+\mathrm{q})\mathrm{Re}\left({\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{t}}^2{\mathrm{e}}^{\mathrm{i}({\mathrm{p}}^2+\mathrm{pq})\mathrm{t}+\mathrm{q}}}{{\mathrm{t}}^2+1}\mathrm{dt}\right)$$$$\phi \left(\mathrm{z}\right)=\frac{{\mathrm{z}}^2{\mathrm{e}}^{\mathrm{i}(({\mathrm{p}}^2+\mathrm{pq})\mathrm{t}+\mathrm{q})}}{{\mathrm{z}}^2+1}\mathrm{but}{\lim }_{\mathrm{z}\to \mathrm{\infty }}\mid \mathrm{z}\phi \left(\mathrm{z}\right)\mid \nRightarrow 0\mathrm{this}\mathrm{integral}\mathrm{is}\mathrm{not}$$$$\mathrm{convergent}....!$$

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