Integration Questions

Question Number 117213 by bobhans last updated on 10/Oct/20

$$\:\:\underset{\mathrm{0}} {\overset{\:\:\:\:\:\:\:\infty} {\int}}\:\frac{\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{4}}\right)−\mathrm{tan}^{−\mathrm{1}} \left(\frac{\mathrm{x}}{\mathrm{6}}\right)}{\mathrm{x}}\:\mathrm{dx}\:=? \\$$

Answered by TANMAY PANACEA last updated on 10/Oct/20

$${taking}\:{help}\:{from}\:{books} \\$$ $${Frullani}\:{intregal} \\$$ $$\int_{\mathrm{0}} ^{\infty} \frac{{f}\left({ax}\right)−{f}\left({bx}\right)}{{x}}{dx}=\left({A}−{B}\right){ln}\left(\frac{{b}}{{a}}\right) \\$$ $${A}=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{f}\left({x}\right)\:\:\:\:{B}=\underset{{x}\rightarrow\infty} {\mathrm{lim}}{f}\left({x}\right) \\$$ $${here}\:{f}\left({x}\right)={tan}^{−\mathrm{1}} \left({x}\right) \\$$ $$\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}{tan}^{−\mathrm{1}} \left({x}\right)=\mathrm{0}={A} \\$$ $$\underset{{x}\rightarrow\infty} {\mathrm{lim}}\:{tan}^{−\mathrm{1}} \left({x}\right)={tan}^{−\mathrm{1}} \left(\infty\right)=\frac{\pi}{\mathrm{2}}={B} \\$$ $${a}=\frac{\mathrm{1}}{\mathrm{4}}\:\:\:{b}=\frac{\mathrm{1}}{\mathrm{6}} \\$$ $${answer}\:=\left(\mathrm{0}−\frac{\pi}{\mathrm{2}}\right){ln}\left(\frac{\mathrm{1}×\mathrm{4}}{\mathrm{6}×\mathrm{1}}\right)=\frac{\pi}{\mathrm{2}}{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\$$

Commented bybobhans last updated on 10/Oct/20

$$\mathrm{waw}....\mathrm{do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{that}\:\mathrm{book}? \\$$

Commented byTANMAY PANACEA last updated on 10/Oct/20

$${yes}\:{sir}...{hard}\:{copy}...{i}\:{do}\:{not}\:{know}\:{how}\:{to}\:{share}\:{image} \\$$ $${or}\:{documents}\:{in}\:{pdf}\:{form} \\$$

Commented bybemath last updated on 10/Oct/20

$${may}\:{be}\:{sir}\:{can}\:{sent}\:{by}\:{gmail} \\$$

Answered by AbduraufKodiriy last updated on 10/Oct/20

$$\boldsymbol{{I}}=\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}}{\mathrm{4}}\right)−\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}}{\mathrm{6}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{arctan}}\left(\frac{\mathrm{2}\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{24}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}} \\$$ $$\boldsymbol{{I}}\left(\boldsymbol{{t}}\right)=\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{arctan}}\left(\frac{\mathrm{2}\boldsymbol{{tx}}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{24}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}}\:\Rightarrow\:\boldsymbol{{I}}\left(\mathrm{0}\right)=\mathrm{0} \\$$ $$\boldsymbol{{I}}\:'\left(\boldsymbol{{t}}\right)=\int_{\mathrm{0}} ^{\:\infty} \partial_{\boldsymbol{{t}}} \frac{\boldsymbol{{arctan}}\left(\frac{\mathrm{2}\boldsymbol{{tx}}}{\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{24}}\right)}{\boldsymbol{{x}}}\boldsymbol{{dx}}=\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{2}\left(\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{24}\right)}{\boldsymbol{{x}}^{\mathrm{4}} +\mathrm{48}\boldsymbol{{x}}^{\mathrm{2}} +\mathrm{576}+\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} \boldsymbol{{x}}^{\mathrm{2}} }\boldsymbol{{dx}}= \\$$ $$=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{\mathrm{1}+\frac{\mathrm{24}}{\boldsymbol{{x}}^{\mathrm{2}} }}{\boldsymbol{{x}}^{\mathrm{2}} +\frac{\mathrm{576}}{\boldsymbol{{x}}^{\mathrm{2}} }+\mathrm{48}+\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} }\boldsymbol{{dx}}=\mathrm{2}\int_{\mathrm{0}} ^{\:\infty} \frac{\boldsymbol{{d}}\left(\boldsymbol{{x}}−\frac{\mathrm{24}}{\boldsymbol{{x}}}\right)}{\left(\boldsymbol{{x}}−\frac{\mathrm{24}}{\boldsymbol{{x}}}\right)^{\mathrm{2}} +\mathrm{96}+\mathrm{4}\boldsymbol{{t}}^{\mathrm{2}} }= \\$$ $$=\frac{\mathrm{2}}{\mathrm{2}\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}−\frac{\mathrm{24}}{\boldsymbol{{x}}}}{\mathrm{2}\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\right)\mid_{\mathrm{0}} ^{\infty} +\boldsymbol{{C}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\boldsymbol{{arctan}}\left(\frac{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{24}}{\mathrm{2}\boldsymbol{{x}}\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\right)\mid_{\mathrm{0}} ^{\infty} = \\$$ $$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\left(\frac{\pi}{\mathrm{2}}−\left(−\frac{\pi}{\mathrm{2}}\right)\right)=\frac{\pi}{\:\sqrt{\mathrm{24}+\boldsymbol{{t}}^{\mathrm{2}} }}\:\Rightarrow \\$$ $$\Rightarrow\:\boldsymbol{{I}}\left(\boldsymbol{{t}}\right)=\pi\boldsymbol{{ln}}\left(\boldsymbol{{t}}+\sqrt{\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{24}}\right)+\boldsymbol{{C}} \\$$ $$\boldsymbol{{I}}\left(\mathrm{0}\right)=\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{24}+\boldsymbol{{C}}=\mathrm{0}\:\Rightarrow\:\boldsymbol{{C}}=−\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{24} \\$$ $$\boldsymbol{{So}},\:\boldsymbol{{I}}\left(\boldsymbol{{t}}\right)=\pi\boldsymbol{{ln}}\left(\boldsymbol{{t}}+\sqrt{\boldsymbol{{t}}^{\mathrm{2}} +\mathrm{24}}\right)−\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{24} \\$$ $$\boldsymbol{{I}}=\boldsymbol{{I}}\left(\mathrm{1}\right)=\pi\boldsymbol{{ln}}\left(\mathrm{1}+\mathrm{5}\right)−\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\mathrm{24}=\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\frac{\mathrm{36}}{\mathrm{24}}=\frac{\pi}{\mathrm{2}}\boldsymbol{{ln}}\frac{\mathrm{3}}{\mathrm{2}} \\$$

Answered by mathmax by abdo last updated on 10/Oct/20

$$\mathrm{I}\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\mathrm{ax}\right)−\mathrm{arctan}\left(\mathrm{bx}\right)}{\mathrm{x}}\mathrm{dx}\:=\mathrm{limI}\left(\xi\right)\mathrm{with}\:\:\mathrm{I}\left(\xi\right)=\int_{\mathrm{0}} ^{\xi} \:\frac{\mathrm{arctan}\left(\mathrm{ax}\right)−\mathrm{arctan}\left(\mathrm{bx}\right)}{\mathrm{x}}\mathrm{dx}=\mathrm{I} \\$$ $$\int_{\mathrm{0}} ^{\infty} \left(...\right)\mathrm{dx}\:=\mathrm{lim}_{\xi\rightarrow+\infty} \mathrm{I}\left(\xi\right) \\$$ $$\mathrm{I}\left(\xi\right)\:=\int_{\mathrm{0}} ^{\xi} \:\frac{\mathrm{arctan}\left(\mathrm{ax}\right)}{\mathrm{x}}\mathrm{dx}\left(\rightarrow\mathrm{ax}\:=\mathrm{u}\right)−\int_{\mathrm{0}} ^{\xi} \:\frac{\mathrm{arctan}\left(\mathrm{bx}\right)}{\mathrm{x}}\mathrm{dx}\left(\mathrm{bx}=\mathrm{v}\right) \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{a}\xi} \:\frac{\mathrm{arctanu}}{\mathrm{u}}\mathrm{du}\:−\int_{\mathrm{0}} ^{\mathrm{b}\xi} \:\frac{\mathrm{arctanv}}{\mathrm{v}}\mathrm{dv}\:=\int_{\mathrm{b}\xi} ^{\mathrm{a}\xi} \:\frac{\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{u}}\mathrm{du} \\$$ $$\left.\exists\:\mathrm{c}\:\in\right]\mathrm{b}\xi\:,\mathrm{a}\xi\left[\:\:/\:\int_{\mathrm{b}\xi} ^{\mathrm{a}\xi} \:\frac{\mathrm{arctan}\left(\mathrm{u}\right)}{\mathrm{u}}\mathrm{du}\:=\mathrm{arctanc}\:\int_{\mathrm{b}\xi} ^{\mathrm{a}\xi} \:\frac{\mathrm{du}}{\mathrm{u}}\:=\mathrm{arctancln}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\right. \\$$ $$\Rightarrow\mathrm{lim}_{\xi\rightarrow+\infty} \mathrm{I}\left(\xi\right)\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{a}}{\mathrm{b}}\right)\:\:\mathrm{with}\:\mathrm{a}=\frac{\mathrm{1}}{\mathrm{4}}\:\mathrm{and}\:\mathrm{b}=\frac{\mathrm{1}}{\mathrm{6}}\:\Rightarrow\frac{\mathrm{a}}{\mathrm{b}}=\frac{\mathrm{6}}{\mathrm{4}}=\frac{\mathrm{3}}{\mathrm{2}} \\$$ $$\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{4}}\right)−\mathrm{arctan}\left(\frac{\mathrm{x}}{\mathrm{6}}\right)}{\mathrm{x}}\mathrm{dx}\:=\frac{\pi}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{3}}{\mathrm{2}}\right) \\$$