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Question Number 208318 by SANOGO last updated on 11/Jun/24

$$calcullimn\to +\mathrm{\infty }$$$${\int }_0^{+\mathrm{\infty }}\frac{cos\left(nx\right)}{(nx+1)(1+x^2)}dx$$

Answered by Berbere last updated on 11/Jun/24

$$\mid {\int }_0^{+\mathrm{\infty }}\frac{cos\left(nx\right)}{(1+nx)(1+x^2)}dx\mid ⩽{\int }_0^{\mathrm{\infty }}\frac{\mid cos\left(nx\right)\mid }{(1+nx)(1+x^2)}dx⩽{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+nx)(1+x^2)}$$$$x\to \frac{1}{x^2}\Rightarrow {\int }_0^{\mathrm{\infty }}\frac{xdx}{(1+x^2)(n+x)}={\int }_0^{\mathrm{\infty }}\frac{-n}{1+n^2}.\frac{1}{(n+x)}+\frac{\frac{nx}{1+n^2}+\frac{1}{n^2+1}}{(1+x^2)}dx$$$$\frac{1}{n^2+1}{\int }_0^{\mathrm{\infty }}-\frac{n}{n+x}+\frac{nx+1}{x^2+1}dx=\frac{1}{n^2+1}{[-nln(n+x)+\frac{n}{2}ln(x^2+1)+{\tan }^{-1}\left(x\right)]}_0^{\mathrm{\infty }}$$$$=\frac{n}{n^2+1}{\left[ln\left(\frac{\sqrt{x^2+1}}{x+n}\right)\right]}_0^{\mathrm{\infty }}+\frac{\pi }{2(n^2+1)}$$$$=-\frac{nln\left(n\right)}{n^2+1}+\frac{\pi }{2(n^2+1)}$$$$\mid {\int }_0^{\mathrm{\infty }}\frac{cos\left(nx\right)}{(nx+1)(1+x^2)}dx\mid ⩽\frac{1}{n^2+1}(\frac{\pi }{2}-nln\left(n\right))=\frac{\pi }{2(n^2+1)}-\frac{\frac{ln\left(n\right)}{n}}{1+\frac{1}{n^2}}\stackrel{n\to \mathrm{\infty }}{\to }0$$$$$$

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