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Question Number 116318 by Bird last updated on 03/Oct/20

$$1)explicitef\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan(a+x)}{x^2+4}dx$$$$1)1)calculate{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan(1+x)}{x^2+4}dx$$$$and{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{arctan(3+x)}{x^2+4}dx$$

Answered by Olaf last updated on 03/Oct/20

$$1)$$$$\mathrm{F}\left(x\right)={\int }_a^{b}f(x,t)dt\Rightarrow {\mathrm{F}}^{\prime }\left(x\right)={\int }_a^b\frac{\partial f}{\partial x}(x,t)dt$$$$\mathrm{Here}\mathrm{we}\mathrm{have}:$$$$f^{\prime }\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{1}{x^2+4}\times \frac{1}{{(a+x)}^2+1}dx$$$$f^{\prime }\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}[\frac{\mathrm{A}x+\mathrm{B}}{x^2+4}+\frac{\mathrm{C}(x+a)+\mathrm{D}}{{(x+a)}^2+1}]dx$$$$\mathrm{A}=-\frac{2a}{(a^2+1)(a^2+9)}$$$$\mathrm{B}=\frac{a^2-3}{(a^2+1)(a^2+9)}$$$$\mathrm{C}=\frac{2a}{(a^2+1)(a^2+9)}=-\mathrm{A}$$$$\mathrm{D}=\frac{a^2+3}{(a^2+1)(a^2+9)}$$$$f^{\prime }\left(a\right)=\frac{\mathrm{A}}{2}{[\ln \mid \frac{x^2+4}{{(x+a)}^2+1}\mid ]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}$$$$+\frac{\mathrm{B}}{2}{\left[\arctan \left(\frac{x}{2}\right)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}+\mathrm{D}{\left[\arctan (x+a)\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}$$$$f^{\prime }\left(a\right)=\frac{\pi }{2}\mathrm{B}+\pi \mathrm{D}$$$$f^{\prime }\left(a\right)=\frac{3\pi }{2}.\frac{1}{a^2+9}$$$$\Rightarrow f\left(a\right)=\frac{\pi }{2}\arctan \frac{a}{3}+\mathrm{K}$$$$\underset{a\to +\mathrm{\infty }}{\lim }f\left(a\right)=\frac{{\pi }^2}{4}+\mathrm{K}$$$$\mathrm{and}\underset{a\to +\mathrm{\infty }}{\lim }f\left(a\right)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\pi }{2}.\frac{1}{x^2+4}dx$$$$\underset{a\to +\mathrm{\infty }}{\lim }f\left(a\right)=\frac{\pi }{4}{\left[\arctan \frac{x}{2}\right]}_{-\mathrm{\infty }}^{+\mathrm{\infty }}=\frac{{\pi }^2}{4}$$$$\Rightarrow \frac{{\pi }^2}{4}+\mathrm{K}=\frac{{\pi }^2}{4},\mathrm{K}=0$$$$\mathrm{Finally}f\left(a\right)=\frac{\pi }{2}\arctan \frac{a}{3}$$$$1)1)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan (1+x)}{x^2+4}dx=f\left(1\right)=\frac{\pi }{2}\arctan \frac{1}{3}$$$$(\arctan \frac{1}{3}=0,322...)$$$$$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan (3+x)}{x^2+4}dx=f\left(3\right)=\frac{\pi }{2}\arctan \left(1\right)$$$${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\arctan (3+x)}{x^2+4}dx=f\left(3\right)=\frac{{\pi }^2}{8}$$$$$$$$\mathbf{Please}\mathbf{verify}\mathbf{my}\mathbf{calculous}...$$$$$$

Commented by mathmax by abdo last updated on 03/Oct/20

$$\mathrm{thank}\mathrm{you}\mathrm{sir}.$$

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