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Question Number 32734 by caravan msup abdo. last updated on 01/Apr/18

$$1)a⩾0calculate{\int }_0^a\frac{n^2-x^2}{{(n^2+x^2)}^2}dxwith$$$$nintegr$$$$2)find{\int }_0^{\mathrm{\infty }}\frac{n^2-x^2}{{(n^2+x^2)}^2}dx$$$$3)calculate\sum _{n=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{n^2-x^2}{{(n^2+x^2)}^2}dx.$$

Commented by abdo imad last updated on 08/Apr/18

$$letput{I}_n\left(a\right)={\int }_0^a\frac{n^2-x^2}{(n^2+x^2)}dx$$$$I_n\left(a\right)=n^2{\int }_0^a\frac{dx}{{(n^2+x^2)}^2}-{\int }_0^a\frac{n^2+x^2-n^2}{(n^2+x^2)}dx$$$$=2n^2{\int }_0^a\frac{dx}{{(n^2+x^2)}^2}-{\int }_0^a\frac{dx}{n^2+x^2}.ch.x=ntgive$$$${\int }_0^a\frac{dx}{n^2+x^2}={\int }_0^{\frac{a}{n}}\frac{ndt}{n^2(1+t^2)}=\frac{1}{n}arctan\left(\frac{a}{n}\right)also$$$${\int }_0^a\frac{dx}{{(n^2+x^2)}^2}={\int }_0^{\frac{a}{n}}\frac{ndt}{n^4{(1+t^2)}^2}=\frac{1}{n^3}{\int }_0^{\frac{a}{n}}\frac{dt}{{(1+t^2)}^2}$$$${=}_{t=tan\theta }\frac{1}{n^3}{\int }_0^{arctan\left(\frac{a}{n}\right)}\frac{(1+{tan}^2\theta )d\theta }{{(1+{tan}^2\theta )}^2}$$$$=\frac{1}{n^3}{\int }_0^{arctan\left(\frac{a}{n}\right)}\frac{d\theta }{1+{tan}^2\theta }=\frac{1}{n^3}{\int }_0^{arctan\left(\frac{a}{n}\right)}\left({cos}^2\theta \right)d\theta$$$$=\frac{1}{2n^3}{\int }_0^{arctan\left(\frac{a}{n}\right)}(1+cos\left(2\theta \right))d\theta$$$$=\frac{arctan\left(\frac{a}{n}\right)}{2n^3}+\frac{1}{4n^3}{\left[sin\left(2\theta \right)\right]}_0^{arctan\left(\frac{a}{n}\right)}$$$$=\frac{arctan\left(\frac{a}{n}\right)}{2n^3}+\frac{1}{4n^3}sin\left(2arctan\left(\frac{a}{n}\right)\right)\Rightarrow$$$$I_n\left(a\right)=\frac{arctan\left(\frac{a}{n}\right)}{n}+\frac{1}{2n}sin\left(2arctan\left(\frac{a}{n}\right)\right)-\frac{1}{n}arctan\left(\frac{a}{n}\right)$$$$I_n\left(a\right)=\frac{1}{2n}sin\left(2arctan\left(\frac{a}{n}\right)\right).$$

Commented by abdo imad last updated on 08/Apr/18

$$wehavesin\left(2arctan\left(\frac{a}{n}\right)\right)=2sin\left(arctan\left(\frac{a}{n}\right)\right)cos\left(arctsn\left(\frac{a}{n}\right)\right)$$$$butwehavethefrmulaesin\left(arctanx\right)=\frac{x}{\sqrt{1+x^2}}$$$$cos\left(arctanx\right)=\frac{1}{\sqrt{1+x^2}}\Rightarrow$$$$I_n\left(a\right)=\frac{1}{n}\frac{\frac{a}{n}}{\sqrt{1+\frac{a^2}{n^2}}}\frac{1}{\sqrt{1+\frac{a^2}{n^2}}}=\frac{a}{(1+\frac{a^2}{n^2})}=\frac{{an}^2}{n^2+a^2}$$

Commented by abdo imad last updated on 08/Apr/18

$$2){\int }_0^{\mathrm{\infty }}\frac{n^2-x^2}{{(n^2+x^2)}^2}={lim}_{a\to +\mathrm{\infty }}{I}_n\left(a\right)={lim}_{a\to +\mathrm{\infty }}\frac{{an}^2}{n^2+a^2}$$$$={lim}_{a\to +\mathrm{\infty }}\frac{n^2}{a}=0.$$

Answered by hknkrc46 last updated on 04/Apr/18

$$1)\mathrm{x}=\mathrm{ntanu}\Rightarrow \mathrm{dx}=\mathrm{n}(1+{\tan }^2\mathrm{u})\mathrm{du}={\mathrm{nsec}}^2\mathrm{udu}$$$$\mathrm{x}\to \mathrm{a},\mathrm{u}\to \arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)\wedge \mathrm{x}\to 0,\mathrm{u}\to 0$$$${\int }_0^{\mathrm{a}}\frac{{\mathrm{n}}^2-{\mathrm{x}}^2}{{({\mathrm{n}}^2+{\mathrm{x}}^2)}^2}\mathrm{dx}={\int }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}\frac{{\mathrm{n}}^2-{\left(\mathrm{ntanu}\right)}^2}{{({\mathrm{n}}^2+{\left(\mathrm{ntanu}\right)}^2)}^2}{\mathrm{nsec}}^2\mathrm{udu}$$$$={\int }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}\frac{{\mathrm{n}}^2-{\mathrm{n}}^2{\tan }^2\mathrm{u}}{{({\mathrm{n}}^2+{\mathrm{n}}^2{\tan }^2\mathrm{u})}^2}{\mathrm{nsec}}^2\mathrm{udu}$$$$={\int }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}\frac{{\mathrm{n}}^2(1-{\tan }^2\mathrm{u})}{{\left({\mathrm{n}}^2(1+{\tan }^2\mathrm{u})\right)}^2}{\mathrm{nsec}}^2\mathrm{udu}$$$$={\int }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}\frac{\frac{{\mathrm{n}}^2\cos 2\mathrm{u}}{{\cos }^2\mathrm{u}}}{{\mathrm{n}}^4{\sec }^4\mathrm{u}}{\mathrm{nsec}}^2\mathrm{udu}$$$$={\int }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}\frac{\cos 2\mathrm{u}}{\mathrm{n}}\mathrm{du}=\frac{\sin 2\mathrm{u}}{2\mathrm{n}}{\mid }_0^{\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)}=\frac{\sin \left(2\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)\right)}{2\mathrm{n}}$$$$=\frac{\sin \left(\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)\right)\cos \left(\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)\right)}{\mathrm{n}}=\frac{\mathrm{a}}{{\mathrm{a}}^2+{\mathrm{n}}^2}$$$$★\arctan \left(\frac{\mathrm{a}}{\mathrm{n}}\right)=\psi \Rightarrow \tan \psi =\frac{\mathrm{a}}{\mathrm{n}}\Rightarrow \sin \psi =\frac{\mathrm{a}}{\sqrt{{\mathrm{a}}^2+{\mathrm{n}}^2}}\wedge \cos \psi =\frac{\mathrm{n}}{\sqrt{{\mathrm{a}}^2+{\mathrm{n}}^2}}$$$$2){\int }_0^{\mathrm{\infty }}\frac{{\mathrm{n}}^2-{\mathrm{x}}^2}{{({\mathrm{n}}^2+{\mathrm{x}}^2)}^2}\mathrm{dx}={\lim }_{\mathrm{a}\to \mathrm{\infty }}{\int }_0^{\mathrm{a}}\frac{{\mathrm{n}}^2-{\mathrm{x}}^2}{{({\mathrm{n}}^2+{\mathrm{x}}^2)}^2}\mathrm{dx}={\lim }_{\mathrm{a}\to \mathrm{\infty }}\frac{\mathrm{a}}{{\mathrm{a}}^2+{\mathrm{n}}^2}=0$$$$3)\sum _{\mathrm{n}=1}^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{n}}^2-{\mathrm{x}}^2}{{({\mathrm{n}}^2+{\mathrm{x}}^2)}^2}\mathrm{dx}={\lim }_{\mathrm{k}\to \mathrm{\infty }}\sum _{\mathrm{n}=1}^{\mathrm{k}}{\int }_0^{\mathrm{\infty }}\frac{{\mathrm{n}}^2-{\mathrm{x}}^2}{{({\mathrm{n}}^2+{\mathrm{x}}^2)}^2}\mathrm{dx}=0$$$$$$

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