Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 116097 by mathmax by abdo last updated on 30/Sep/20

$$\mathrm{calculate}{\int }_0^{\mathrm{\infty }}\frac{\mathrm{lnx}}{{\mathrm{x}}^4+1}\mathrm{dx}$$

Answered by mnjuly1970 last updated on 01/Oct/20

$$solution:\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{ln\left(x\right)}{1+x^4}=???$$$$x^4=t\Rightarrow x=t^{\frac{1}{4}}\Rightarrow dx=\frac{1}{4}{t}^{\frac{-3}{4}}$$$$\mathrm{\Omega }=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{ln\left(t\right)t^{-\frac{3}{4}}}{1+t}dt$$$$f\left(\lambda \right)={\int }_0^{\mathrm{\infty }}\frac{t^{-\lambda }}{1+t}dt\Rightarrow \mathrm{\Omega }=\frac{-1}{16}f{}^{\prime }\left(\frac{3}{4}\right)$$$$but::f\left(\lambda \right)=\mathrm{\Gamma }\left(\lambda \right)\mathrm{\Gamma }(1-\lambda );0<\lambda <1$$$$f\left(\lambda \right)=\frac{\pi }{sin\left(\lambda \pi \right)}\Rightarrow f{}^{\prime }\left(\lambda \right)={[-\frac{{\pi }^{2}cos\left(\pi \lambda \right)}{{sin}^2\left(\pi \lambda \right)}]}_{\lambda =\frac{3}{4}}$$$$f{}^{\prime }\left(\frac{3}{4}\right)={\pi }^2\ast \frac{\frac{\sqrt{2}}{2}}{\frac{1}{2}}={\pi }^2\sqrt{2}$$$$\mathrm{\Omega }:=-\frac{1}{16}f{}^{\prime }\left(\frac{3}{4}\right)=-\frac{{\pi }^2\sqrt{2}}{16}✓$$$$...m.n.july.1970...$$$$$$$$$$$$$$$$$$$$$$

Commented by 1549442205PVT last updated on 01/Oct/20

$$\mathrm{third}\mathrm{line}\stackrel{?}{\to }\mathrm{fourth}\mathrm{line}(\mathrm{lnt}\mathrm{missed}?)$$

Commented by mnjuly1970 last updated on 01/Oct/20

$$method:{feynman}^{\prime }strick$$$$pleaselookatagain.$$$$$$$$$$

Commented by 1549442205PVT last updated on 01/Oct/20

$$\mathrm{Thank}\mathrm{Sir},\mathrm{i}\mathrm{understanded}.{\left({\mathrm{a}}^{\mathrm{x}}\right)}^{\prime }={\mathrm{a}}^{\mathrm{x}}\mathrm{lna}$$$$\mathrm{and}\mathrm{put}\mathrm{t}=\frac{1}{\mathrm{x}}-1\Rightarrow \mathrm{f}\left(\lambda \right)={\int }_0^1{\mathrm{x}}^{-\lambda }{(1-\mathrm{x})}^{1-\lambda }=\beta (\lambda ,1-\lambda )=\mathrm{\Gamma }\left(\lambda \right)\mathrm{\Gamma }(1-\lambda )=\frac{\pi }{\sin \left(\lambda \pi \right)}$$$$\mathrm{Great}\mathrm{Sir}\mathrm{thanh}\mathrm{you}\mathrm{very}\mathrm{much}$$

Commented by mnjuly1970 last updated on 01/Oct/20

$$youarewelcome.$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{grest}\mathrm{sir}$$

Answered by mathdave last updated on 01/Oct/20

$$solution$$$$letI={\int }_0^1\frac{\ln x}{x^4+1}dx+{\int }_1^{\mathrm{\infty }}\frac{\ln x}{1+x^4}dx=A+B$$$$puttingx=\frac{1}{x}intoB$$$$I={\int }_0^1\frac{\ln x}{1+x^4}dx+{\int }_1^0\frac{\ln \left(\frac{1}{x}\right)}{1+\frac{1}{x^4}}\times -\frac{1}{x^2}dx={\int }_0^1\frac{\ln x}{1+x^4}dx-{\int }_0^1\frac{x^2\ln x}{1+x^4}dx$$$$I={(-)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^{4n}\ln xdx-{(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}{\int }_0^1{x}^2.x^{4n}\ln xdx$$$$I={(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{4n}.x^{a-1}dx-{(-1)}^n\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}{\int }_0^1{x}^{4n+2}.x^{a-1}dx$$$$I=\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\frac{{(-1)}^n}{(4n+a)}-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{\partial }{\partial a}{\mid }_{a=1}\frac{{(-1)}^n}{(4n+2+a)}$$$$I=-\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(4n+1)}^2}+\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(4n+3)}^2}=-\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(4n+1)}^2}$$$$note\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^n}{{(ax+1)}^n}=-\frac{\pi }{a^n}\underset{z\to -\frac{1}{a}}{\lim }\frac{1}{(n-1)!}\bullet \frac{d^{n-1}}{{dz}^{n-1}}\left(\mathrm{cosec}\left(\pi z\right)\right)$$$$I=-\underset{n=-\mathrm{\infty }}{\sum ^{\mathrm{\infty }}}\frac{1}{{(4n+1)}^2}=-(-\frac{\pi }{4^2})\underset{z\to -\frac{1}{4}}{\lim }\frac{1}{(2-1)!}\bullet \frac{d}{dz}\left(cosec\left(\pi z\right)\right)$$$$I=\frac{\pi }{16}\underset{z\to -\frac{1}{4}}{\lim }\frac{1}{1!}(-\pi \mathrm{cosec}\left(\pi z\right)\cot \left(\pi z\right))$$$$I=-\frac{{\pi }^2}{16}\mathrm{cosec}(-\frac{\pi }{4})\cot (-\frac{\pi }{4})=-\frac{{\pi }^2}{16}(-\sqrt{2})(-1)=-\frac{{\pi }^2}{16}\sqrt{2}$$$$\because {\int }_0^{\mathrm{\infty }}\frac{\ln x}{1+x^4}dx=-\frac{{\pi }^2}{8\sqrt{2}}$$$$bymathdave\left(01/10/2020\right)$$

Commented by Tawa11 last updated on 06/Sep/21

$$\mathrm{great}\mathrm{sir}$$

Answered by Bird last updated on 01/Oct/20

$$letf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{1+t}dtwith0<a<1$$$$wehavef\left(a\right)=\frac{\pi }{sin\left(\pi a\right)}$$$$andf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{e^{(a-1)lnt}}{1+t}dt\Rightarrow$$$$f^{{}^{\prime }}\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{lnt{t}^{a-1}}{1+t}dt={\int }_0^{\mathrm{\infty }}\frac{t^{a-1}lnt}{1+t}dt$$$${\int }_0^{\mathrm{\infty }}\frac{lnx}{x^4+1}dx{=}_{x=t^{\frac{1}{4}}}\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{lnt}{1+t}.\frac{1}{4}{t}^{\frac{1}{4}-1}dt$$$$=\frac{1}{16}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{1}{4}-1}lnt}{1+t}dt$$$$=\frac{1}{16}{f}^{{}^{\prime }}\left(\frac{1}{4}\right)but{f}^{{}^{\prime }}\left(a\right)=-\frac{{\pi }^{2}cos\left(\pi a\right)}{{sin}^2\left(\pi a\right)}$$$$\Rightarrow f^{{}^{\prime }}\left(\frac{1}{4}\right)=-{\pi }^2\times \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}}=\frac{-2{\pi }^2}{\sqrt{2}}$$$$=-\sqrt{2}{\pi }^2\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{lnx}{1+x^4}dx=-\frac{{\pi }^2\sqrt{2}}{16}$$

Commented by mnjuly1970 last updated on 01/Oct/20

$$excellentsir..$$

Commented by Bird last updated on 01/Oct/20

$$youarewelcome$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com