Question Number 39030 by maxmathsup by imad last updated on 01/Jul/18
$$\left.\mathrm{1}\right)\:{let}\:{f}\left({x}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} }\:\:{with}\:{x}\:>\mathrm{0} \\ $$$${find}\:{a}\:{simple}\:{form}\:{of}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} } \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+\mathrm{3}{t}^{\mathrm{4}} } \\ $$
Commented by maxmathsup by imad last updated on 02/Jul/18
$$\left.\mathrm{1}\right){we}\:{have}\:\mathrm{2}{f}\left({x}\right)=\:\int_{−\infty} ^{+\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{x}^{\mathrm{2}} {t}^{\mathrm{4}} }\:{let}\:\varphi\left({z}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} {z}^{\mathrm{4}} \:+\mathrm{1}} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({z}^{\mathrm{4}} \:+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left({z}^{\mathrm{2}} \:−\frac{{i}}{{x}}\right)\left({z}^{\mathrm{2}} \:+\frac{{i}}{{x}}\right)} \\ $$$$=\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} \left(\:{z}\:−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\left({z}+\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{\frac{{i}\pi}{\mathrm{4}}} \right)\:+{Res}\left(\varphi,−\frac{\mathrm{1}}{\:\sqrt{{x}}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right)\:{but}\right. \\ $$$${Res}\left(\varphi,{z}_{{i}} \right)\:=\:\frac{\mathrm{1}}{\mathrm{4}{x}^{\mathrm{2}} {z}_{{i}} ^{\mathrm{3}} }\:\:{wehave}\:{z}_{{i}} ^{\mathrm{4}} \:=\frac{−\mathrm{1}}{{x}^{\mathrm{2}} }\:\Rightarrow{Res}\left(\varphi,{z}_{{i}} \right)=\:\frac{{z}_{{i}} }{\mathrm{4}{x}^{\mathrm{2}} \:\left(−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)}\:=−\frac{{z}_{{i}} }{\mathrm{4}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{−\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}}\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:\:\:\:\:+\frac{\mathrm{1}}{\mathrm{4}\sqrt{{x}}}\:{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\} \\ $$$$=−\frac{{i}\pi}{\mathrm{2}\sqrt{{x}}}\left\{\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \:−{e}^{−\frac{{i}\pi}{\mathrm{4}}} \right\}\:=−\frac{{i}\pi}{\mathrm{2}\sqrt{{x}}}\left\{\mathrm{2}{i}\:\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right\}\:=\frac{\pi\sqrt{\mathrm{2}}}{\:\sqrt{{x}}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\:\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:. \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:=\:{f}\left(\mathrm{1}\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dt}}{\mathrm{1}+\mathrm{3}{t}^{\mathrm{4}} }\:\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)}\:. \\ $$
Commented by ajfour last updated on 02/Jul/18
$${Sir},\:{have}\:{you}\:{found}\:\:\mathrm{2}{f}\left({x}\right)\:{or}\:{f}\left({x}\right); \\ $$$${please}\:{check}.. \\ $$
Commented by maxmathsup by imad last updated on 02/Jul/18
$${error}\:{from}\:{line}\:\mathrm{7}\:\:\:\:\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}\sqrt{{x}}}\:\:\Rightarrow{f}\left({x}\right)=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}\sqrt{{x}}} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{4}} }\:\:={f}\left(\mathrm{1}\right)\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$$\left.\mathrm{3}\right)\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+\mathrm{3}{t}^{\mathrm{4}} }\:\:={f}\left(\sqrt{\mathrm{3}}\right)\:=\:\:\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{4}\left(^{\mathrm{4}} \sqrt{\mathrm{3}}\right)}\:. \\ $$
Answered by behi83417@gmail.com last updated on 02/Jul/18
![(1−ixt^2 )(1+ixt^2 )=0 ⇒t^2 =(1/(ix)),−(1/(ix))⇒t=±(1/( (√(ix)))),((±i)/( (√(ix)))),(1/( (√(ix))))=m m=(1/( (√(ix))))=((√i)/(i(√x)))=−((ie^((iπ)/4) )/( (√x)))=−((i(((√2)/2)+i((√2)/2)))/( (√x)))=((1−i)/( (√(2x)))) I=∫(dt/((t−m)(t+m)(t−im)(t+im)))= (1/((t−m)(t+m)(t−im)(t+im)))= =(a/(t−m))+(b/(t+m))+(c/(t−im))+(d/(t+im)) a=(1/(2m.m^2 (1−i^2 )))=(1/(4m^3 )) b=(1/(−2m.m^2 (1−i^2 )))=((−1)/(4m^3 )) c=(1/(m(i−1)m(i+1)(2im)))=(i/(4m^3 )) d=(1/(−m(i+1)m(i−1)(2im)))=((−i)/(4m^3 )) I=(1/(4m^3 ))∫[(1/(t−m))−(1/(t+m))+(i/(t−im))−(i/(t+im))]dt =(1/(4m^3 ))[ln((t−m)/(t+m))+i.ln((t−im)/(t+im))]= =(1/(4m^3 ))[ln((t^2 −2mt+m^2 )/(t^2 −m^2 ))+i.ln((t^2 −2mit−m^2 )/(t^2 +m^2 ))]= =(1/(4.(((1−i)^3 )/(2x(√(2x))))))[ln((t^2 −2t((1−i)/( (√(2x))))+(i/x))/(t^2 −(i/x)))+i.ln((t^2 −2t((1+i)/( (√(2x))))−(i/x))/(t^2 +(i/x)))]= =−((x(√(2x)))/(4(2i+1)))[ln(((x(√(2x)).t^2 −2(1−i)xt+i.(√(2x)))/(xt^2 −i)))+ +i.ln(((x(√(2x)).t^2 −2(1+i)xt−i(√(2x)))/(xt^2 +i)))]+const. F(x)=F(∞)−F(0)=−((x(√(2x)))/(4(2i+1)))[ln(√(2x))+i.ln(√(2x))− −ln(−(√(2x)))−i.ln(−(√(2x)))]= =((−x(√(2x)))/(4(2i+1)))[ln(−1)+iln(−1)]= =((−x(√(2x)))/(4(−4−1)))(2i−1)(1+i).i𝛑=((−π.x(√(2x)))/(20))(3i+1) F(1)=((−π(√2))/(20))(3i+1) F((√3))=((−π(√(6(√3))))/(20))(3i+1) .](https://www.tinkutara.com/question/Q39046.png)
$$\left(\mathrm{1}−{ixt}^{\mathrm{2}} \right)\left(\mathrm{1}+{ixt}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{2}} =\frac{\mathrm{1}}{{ix}},−\frac{\mathrm{1}}{{ix}}\Rightarrow{t}=\pm\frac{\mathrm{1}}{\:\sqrt{{ix}}},\frac{\pm{i}}{\:\sqrt{{ix}}},\frac{\mathrm{1}}{\:\sqrt{{ix}}}={m} \\ $$$${m}=\frac{\mathrm{1}}{\:\sqrt{{ix}}}=\frac{\sqrt{{i}}}{{i}\sqrt{{x}}}=−\frac{{ie}^{\frac{{i}\pi}{\mathrm{4}}} }{\:\sqrt{{x}}}=−\frac{{i}\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}+{i}\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)}{\:\sqrt{{x}}}=\frac{\mathrm{1}−{i}}{\:\sqrt{\mathrm{2}{x}}} \\ $$$${I}=\int\frac{{dt}}{\left({t}−{m}\right)\left({t}+{m}\right)\left({t}−{im}\right)\left({t}+{im}\right)}= \\ $$$$\frac{\mathrm{1}}{\left({t}−{m}\right)\left({t}+{m}\right)\left({t}−{im}\right)\left({t}+{im}\right)}= \\ $$$$=\frac{{a}}{{t}−{m}}+\frac{{b}}{{t}+{m}}+\frac{{c}}{{t}−{im}}+\frac{{d}}{{t}+{im}} \\ $$$${a}=\frac{\mathrm{1}}{\mathrm{2}{m}.{m}^{\mathrm{2}} \left(\mathrm{1}−{i}^{\mathrm{2}} \right)}=\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{3}} } \\ $$$${b}=\frac{\mathrm{1}}{−\mathrm{2}{m}.{m}^{\mathrm{2}} \left(\mathrm{1}−{i}^{\mathrm{2}} \right)}=\frac{−\mathrm{1}}{\mathrm{4}{m}^{\mathrm{3}} } \\ $$$${c}=\frac{\mathrm{1}}{{m}\left({i}−\mathrm{1}\right){m}\left({i}+\mathrm{1}\right)\left(\mathrm{2}{im}\right)}=\frac{{i}}{\mathrm{4}{m}^{\mathrm{3}} } \\ $$$${d}=\frac{\mathrm{1}}{−{m}\left({i}+\mathrm{1}\right){m}\left({i}−\mathrm{1}\right)\left(\mathrm{2}{im}\right)}=\frac{−{i}}{\mathrm{4}{m}^{\mathrm{3}} } \\ $$$${I}=\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{3}} }\int\left[\frac{\mathrm{1}}{{t}−{m}}−\frac{\mathrm{1}}{{t}+{m}}+\frac{{i}}{{t}−{im}}−\frac{{i}}{{t}+{im}}\right]{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{3}} }\left[{ln}\frac{{t}−{m}}{{t}+{m}}+{i}.{ln}\frac{{t}−{im}}{{t}+{im}}\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{m}^{\mathrm{3}} }\left[{ln}\frac{{t}^{\mathrm{2}} −\mathrm{2}{mt}+{m}^{\mathrm{2}} }{{t}^{\mathrm{2}} −{m}^{\mathrm{2}} }+{i}.{ln}\frac{{t}^{\mathrm{2}} −\mathrm{2}{mit}−{m}^{\mathrm{2}} }{{t}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right]= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}.\frac{\left(\mathrm{1}−{i}\right)^{\mathrm{3}} }{\mathrm{2}{x}\sqrt{\mathrm{2}{x}}}}\left[{ln}\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}\frac{\mathrm{1}−{i}}{\:\sqrt{\mathrm{2}{x}}}+\frac{{i}}{{x}}}{{t}^{\mathrm{2}} −\frac{{i}}{{x}}}+{i}.{ln}\frac{{t}^{\mathrm{2}} −\mathrm{2}{t}\frac{\mathrm{1}+{i}}{\:\sqrt{\mathrm{2}{x}}}−\frac{{i}}{{x}}}{{t}^{\mathrm{2}} +\frac{{i}}{{x}}}\right]= \\ $$$$=−\frac{{x}\sqrt{\mathrm{2}{x}}}{\mathrm{4}\left(\mathrm{2}{i}+\mathrm{1}\right)}\left[\boldsymbol{{ln}}\left(\frac{\boldsymbol{{x}}\sqrt{\mathrm{2}\boldsymbol{{x}}}.\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}−\boldsymbol{{i}}\right)\boldsymbol{{xt}}+\boldsymbol{{i}}.\sqrt{\mathrm{2}\boldsymbol{{x}}}}{\boldsymbol{{xt}}^{\mathrm{2}} −\boldsymbol{{i}}}\right)+\right. \\ $$$$\left.+\boldsymbol{{i}}.\boldsymbol{{ln}}\left(\frac{\boldsymbol{{x}}\sqrt{\mathrm{2}\boldsymbol{{x}}}.\boldsymbol{{t}}^{\mathrm{2}} −\mathrm{2}\left(\mathrm{1}+\boldsymbol{{i}}\right)\boldsymbol{{xt}}−\boldsymbol{{i}}\sqrt{\mathrm{2}\boldsymbol{{x}}}}{\boldsymbol{{xt}}^{\mathrm{2}} +\boldsymbol{{i}}}\right)\right]+\boldsymbol{{const}}. \\ $$$${F}\left({x}\right)={F}\left(\infty\right)−{F}\left(\mathrm{0}\right)=−\frac{{x}\sqrt{\mathrm{2}{x}}}{\mathrm{4}\left(\mathrm{2}{i}+\mathrm{1}\right)}\left[{ln}\sqrt{\mathrm{2}{x}}+\boldsymbol{{i}}.\boldsymbol{{ln}}\sqrt{\mathrm{2}\boldsymbol{{x}}}−\right. \\ $$$$\left.−\boldsymbol{{ln}}\left(−\sqrt{\mathrm{2}\boldsymbol{{x}}}\right)−\boldsymbol{{i}}.\boldsymbol{{ln}}\left(−\sqrt{\mathrm{2}\boldsymbol{{x}}}\right)\right]= \\ $$$$=\frac{−\boldsymbol{{x}}\sqrt{\mathrm{2}\boldsymbol{{x}}}}{\mathrm{4}\left(\mathrm{2}\boldsymbol{{i}}+\mathrm{1}\right)}\left[\boldsymbol{{ln}}\left(−\mathrm{1}\right)+\boldsymbol{{iln}}\left(−\mathrm{1}\right)\right]= \\ $$$$=\frac{−\boldsymbol{{x}}\sqrt{\mathrm{2}\boldsymbol{{x}}}}{\mathrm{4}\left(−\mathrm{4}−\mathrm{1}\right)}\left(\mathrm{2}\boldsymbol{{i}}−\mathrm{1}\right)\left(\mathrm{1}+\boldsymbol{{i}}\right).\boldsymbol{{i}\pi}=\frac{−\pi.{x}\sqrt{\mathrm{2}{x}}}{\mathrm{20}}\left(\mathrm{3}{i}+\mathrm{1}\right) \\ $$$${F}\left(\mathrm{1}\right)=\frac{−\pi\sqrt{\mathrm{2}}}{\mathrm{20}}\left(\mathrm{3}{i}+\mathrm{1}\right) \\ $$$${F}\left(\sqrt{\mathrm{3}}\right)=\frac{−\pi\sqrt{\mathrm{6}\sqrt{\mathrm{3}}}}{\mathrm{20}}\left(\mathrm{3}{i}+\mathrm{1}\right)\:. \\ $$
Answered by ajfour last updated on 02/Jul/18
![f(x)=(1/x^2 )∫_0 ^( ∞) (dt/(t^4 +((1/x))^2 )) let (1/x) = y f(x)=y^2 ∫((dt/t^2 )/(t^2 +(y^2 /t^2 ))) = (y/2)∫(([(1+(y/t^2 ))−(1−(y/t^2 ))]dt)/((t−(y/t))^2 +2y)) =(y/2)[∫(du/(u^2 +((√(2y)))^2 ))−∫^ (dv/(v^2 −((√(2y)))^2 ))] =(y/(2(√(2y))))tan^(−1) ((u/( (√(2y)))))−(y/(4(√(2y))))ln ∣((v−(√(2y)))/(v+(√(2y))))∣+c =(1/(2(√(2x))))tan^(−1) (((t−(1/(tx)))/( (√(2/x)))))∣_0 ^∞ −(1/(4(√(2x))))ln ∣((t+(1/(tx))−(√(2/x)))/(t+(1/(tx))+(√(2/x))))∣_0 ^∞ ⇒ f(x) = (π/(2(√(2x))))](https://www.tinkutara.com/question/Q39075.png)
$${f}\left({x}\right)=\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dt}}{{t}^{\mathrm{4}} +\left(\frac{\mathrm{1}}{{x}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:{let}\:\:\:\frac{\mathrm{1}}{{x}}\:=\:{y} \\ $$$${f}\left({x}\right)={y}^{\mathrm{2}} \int\frac{\frac{{dt}}{{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} +\frac{{y}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}\:=\:\frac{{y}}{\mathrm{2}}\int\frac{\left[\left(\mathrm{1}+\frac{{y}}{{t}^{\mathrm{2}} }\right)−\left(\mathrm{1}−\frac{{y}}{{t}^{\mathrm{2}} }\right)\right]{dt}}{\left({t}−\frac{{y}}{{t}}\right)^{\mathrm{2}} +\mathrm{2}{y}} \\ $$$$\:\:\:\:=\frac{{y}}{\mathrm{2}}\left[\int\frac{{du}}{{u}^{\mathrm{2}} +\left(\sqrt{\mathrm{2}{y}}\right)^{\mathrm{2}} }−\int^{\:} \frac{{dv}}{{v}^{\mathrm{2}} −\left(\sqrt{\mathrm{2}{y}}\right)^{\mathrm{2}} }\right] \\ $$$$\:\:=\frac{{y}}{\mathrm{2}\sqrt{\mathrm{2}{y}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{u}}{\:\sqrt{\mathrm{2}{y}}}\right)−\frac{{y}}{\mathrm{4}\sqrt{\mathrm{2}{y}}}\mathrm{ln}\:\mid\frac{{v}−\sqrt{\mathrm{2}{y}}}{{v}+\sqrt{\mathrm{2}{y}}}\mid+{c} \\ $$$$\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}{x}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}−\frac{\mathrm{1}}{{tx}}}{\:\sqrt{\frac{\mathrm{2}}{{x}}}}\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$\:\:\:\:\:\:\:\:\:\:−\frac{\mathrm{1}}{\mathrm{4}\sqrt{\mathrm{2}{x}}}\mathrm{ln}\:\mid\frac{{t}+\frac{\mathrm{1}}{{tx}}−\sqrt{\frac{\mathrm{2}}{{x}}}}{{t}+\frac{\mathrm{1}}{{tx}}+\sqrt{\frac{\mathrm{2}}{{x}}}}\mid_{\mathrm{0}} ^{\infty} \\ $$$$\:\Rightarrow\:\:\:{f}\left({x}\right)\:=\:\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}{x}}}\: \\ $$
Commented by ajfour last updated on 02/Jul/18
$${wish}\:{i}\:{could}\:{play}\:{with}\:{the}\:{residue} \\ $$$${theorem}\:{even}.. \\ $$
Commented by maxmathsup by imad last updated on 02/Jul/18
$${sir}\:{Ajfour}\:{it}\:{seems}\:{that}\:{you}\:{have}\:{played}\:{a}\:{game}\:{with}\:{this}\:{integral}… \\ $$