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Question Number 31092 by abdo imad last updated on 02/Mar/18

$$find{\int }_0^{\mathrm{\infty }}\frac{ln(1+4x^2)}{1+2x^2}dx.$$

Commented by abdo imad last updated on 07/Mar/18

$$letintroducetheparametricfunction$$$$f\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{ln(1+{tx}^2)}{1+2x^2}dxwitht⩾0wehave$$$$f^{{}^{\prime }}\left(t\right)={\int }_0^{\mathrm{\infty }}\frac{x^2}{(1+{tx}^2)(1+2x^2)}dx$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{2x^2+1-1}{(1+{tx}^2)(1+2x^2)}dx=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{1+{tx}^2}-\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{(1+2x^2)(1+{tx}^2)}but$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+{tx}^2}{=}_{\sqrt{t}x=u}={\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\frac{du}{\sqrt{t}}=\frac{\pi }{2\sqrt{t}}nowletdecompose$$$$F\left(t\right)=\frac{1}{(1+2x^2)(1+{tx}^2)}=\frac{ax+b}{2x^2+1}+\frac{cx+d}{{tx}^2+1}$$$$F(-x)=F\left(x\right)\Rightarrow F\left(x\right)=\frac{-ax+b}{2x^2+1}+\frac{-cx+d}{{tx}^2+1}\Rightarrow a=c=0\Rightarrow$$$$F\left(x\right)=\frac{b}{2x^2+1}+\frac{d}{t^{}{x}^2+1}wehave{lim}_{x\to +\mathrm{\infty }}{x}^{2}F\left(x\right)=0$$$$=\frac{b}{2}+\frac{d}{t}=0\Rightarrow 2d+tb=0\Rightarrow d=-\frac{t}{2}b\Rightarrow$$$$F\left(t\right)=\frac{b}{2x^2+1}-\frac{t}{2}\frac{b}{{tx}^2+1}wehaveF\left(0\right)=1=b-\frac{tb}{2}\Rightarrow$$$$(1-\frac{t}{2})b=1\Rightarrow (2-t)b=2\Rightarrow b=\frac{2}{2-t}\Rightarrow$$$$F\left(x\right)=\frac{2}{(2-t)(2x^2+1)}-\frac{t}{(2-t)({tx}^2+1)}$$$$=\frac{1}{2-t}(\frac{2}{2x^2+1}-\frac{t}{{tx}^2+1})\Rightarrow$$$${\int }_0^{\mathrm{\infty }}F\left(x\right)dx=\frac{2}{2-t}{\int }_0^{\mathrm{\infty }}\frac{dx}{2x^2+1}-\frac{t}{2-t}{\int }_0^{\mathrm{\infty }}\frac{dx}{{tx}^2+1}but$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{2x^2+1}{=}_{\sqrt{2}x=u}{\int }_0^{\mathrm{\infty }}\frac{1}{1+u^2}\frac{du}{\sqrt{2}}=\frac{\pi }{2\sqrt{2}}.$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{tx}^2+1}=\frac{\pi }{2\sqrt{t}}\Rightarrow {\int }_0^{\mathrm{\infty }}F\left(x\right)dx=\frac{\pi }{\sqrt{2}(2-t)}-\frac{t}{2-t}\frac{\pi }{2\sqrt{t}}$$$$=\frac{\pi }{\sqrt{2}(2-t)}-\frac{\pi \sqrt{t}}{2(2-t)}\Rightarrow f^{{}^{\prime }}\left(t\right)=\frac{\pi }{4\sqrt{t}}-\frac{\pi }{2\sqrt{2}(2-t)}+\frac{\pi \sqrt{t}}{4(2-t)}$$$$=\frac{\pi }{4}(\frac{1}{\sqrt{t}}-\frac{2\pi }{\sqrt{2}(2-t)}+\frac{\sqrt{t}}{2-t})\Rightarrow$$$$f\left(t\right)=\frac{\pi }{4}\int \frac{dt}{\sqrt{t}}-\frac{\pi }{2\sqrt{2}}\int \frac{dt}{2-t}+\frac{\pi }{4}\int \frac{\sqrt{t}}{2-t}dt+\lambda$$$$=\frac{\pi }{2}\sqrt{t}+\frac{\pi }{2\sqrt{2}}ln\mid 2-t\mid +\frac{\pi }{4}\int \frac{u}{2-u^2}2udu+\lambda$$$$\frac{\pi }{4}\int \frac{2u^2}{2-u^2}du=\frac{\pi }{2}\int \frac{u^2}{2-u^2}duand$$$$\int \frac{u^2}{2-u^2}du=-\int \frac{2-u^2-2}{2-{\mathit{u}}^2}\mathit{d}\mathit{u}=-\mathit{u}+2\int \frac{du}{2-u^2}$$$$=-u+\frac{1}{\sqrt{2}}\int (\frac{1}{\sqrt{2}-u}+\frac{1}{\sqrt{2}+u})=-\sqrt{t}+\frac{1}{\sqrt{2}}ln\mid \frac{\sqrt{2}+\sqrt{t}}{\sqrt{2}-\sqrt{t}}\mid \Rightarrow$$$$f\left(t\right)=\frac{\pi }{2}\sqrt{t}+\frac{\pi }{2\sqrt{2}}ln\mid 2-t\mid +\frac{\pi }{2}(-\sqrt{t}+\frac{1}{\sqrt{2}}ln\mid \frac{\sqrt{2}+\sqrt{t}}{\sqrt{2}-\sqrt{t}}\mid )+\lambda$$$$f\left(t\right)=\frac{\pi }{2\sqrt{2}}ln\mid 2-t\mid +\frac{\pi }{2\sqrt{2}}ln\mid \frac{\sqrt{2}+\sqrt{t}}{\sqrt{2}-\sqrt{t}}\mid +\lambda wehave$$$$f\left(0\right)=0=\frac{\pi ln2}{2\sqrt{2}}+\lambda \Rightarrow \lambda =-\frac{\pi ln2}{2\sqrt{2}}finally$$$$f\left(t\right)=\frac{\pi }{2\sqrt{2}}ln\mid 2-t\mid +\frac{\pi }{2\sqrt{2}}ln\mid \frac{\sqrt{2}+\sqrt{t}}{\sqrt{2}-\sqrt{t}}\mid -\frac{\pi ln2}{2\sqrt{2}}lettaket=4$$$${\int }_0^{\mathrm{\infty }}\frac{ln(1+4t^2)}{1+2t^2}=\frac{\pi }{2\sqrt{2}}ln2+\frac{\pi }{2\sqrt{2}}ln\mid \frac{\sqrt{2}+2}{\sqrt{2}-2}\mid -\frac{\pi ln2}{2\sqrt{2}}.$$

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