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 85160 by mathmax by abdo last updated on 19/Mar/20

$$1)findf\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^4+a}witha>0$$$$2)findg\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^4+a)}^2}$$$$3)findvalueofintegrals{\int }_0^{\mathrm{\infty }}\frac{dx}{x^4+1},{\int }_0^{\mathrm{\infty }}\frac{dx}{2x^4+8}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^4+1)}^2}and{\int }_0^{\mathrm{\infty }}\frac{dx}{{(2x^4+8)}^2}$$

Commented by mathmax by abdo last updated on 19/Mar/20

$$1)f\left(a\right)={\int }_0^{\mathrm{\infty }}\frac{dx}{x^4+a}{=}_{x{=}^4\sqrt{a}t}{\int }_0^{\mathrm{\infty }}\frac{\left({}^4\sqrt{a}\right)dt}{a(1+t^4)}=\frac{1}{a^{1-\frac{1}{4}}}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}$$$$=\frac{1}{a^{\frac{3}{4}}}{\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}changementt=u^{\frac{1}{4}}give$$$${\int }_0^{\mathrm{\infty }}\frac{dt}{t^4+1}=\frac{1}{4}{\int }_0^{\mathrm{\infty }}\frac{u^{\frac{1}{4}-1}}{1+u}du=\frac{1}{4}\times \frac{\pi }{sin\left(\frac{\pi }{4}\right)}=\frac{\pi }{4\times \frac{\sqrt{2}}{2}}=\frac{\pi }{2\sqrt{2}}\Rightarrow$$$$f\left(a\right)=\frac{\pi }{2\sqrt{2}}{a}^{-\frac{3}{4}}$$$$2)wehave{f}^{{}^{\prime }}\left(a\right)=-{\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^4+a)}^2}=-g\left(a\right)\Rightarrow g\left(a\right)=-f^{{}^{\prime }}\left(a\right)$$$$g\left(a\right)=-\frac{\pi }{2\sqrt{2}}\times \left(\frac{-3}{4}\right)a^{\frac{-3}{4}-1}=\frac{3\pi }{8\sqrt{2}}{a}^{-\frac{7}{4}}$$

Commented by mathmax by abdo last updated on 19/Mar/20

$$3){\int }_0^{\mathrm{\infty }}\frac{dx}{x^4+1}=f\left(1\right)=\frac{\pi }{2\sqrt{2}}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{2x^4+8}=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{dx}{x^4+4}=\frac{1}{2}f\left(4\right)=\frac{\pi }{4\sqrt{2}}{\left(4\right)}^{-\frac{3}{4}}$$$$=\frac{\pi }{4\sqrt{2}{4}^{\frac{3}{4}}}=\frac{\pi }{4^{\frac{7}{4}}\sqrt{2}}=\frac{\pi }{2^{\frac{7}{2}}\sqrt{2}}=\frac{\pi }{8\times 2}=\frac{\pi }{16}$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{{(x^4+1)}^2}=g\left(1\right)=\frac{3\pi \sqrt{2}}{16}$$

Answered by mind is power last updated on 19/Mar/20

$$x=\sqrt{\sqrt{a}tg\left(t\right)}\Rightarrow$$$$f\left(a\right)={\int }_0^{\frac{\pi }{2}}\frac{\sqrt{\sqrt{a}}(1+{tg}^2\left(t\right))}{2\sqrt{tg\left(t\right)}}.\frac{dt}{a(1+{tg}^2\left(t\right))}$$$$=\frac{1}{2a^{\frac{3}{4}}}.{\int }_0^{\frac{\pi }{2}}{cos}^{\frac{1}{2}}\left(t\right){sin}^{\frac{-1}{2}}\left(t\right)dt$$$$=\frac{1}{2a^{\frac{3}{4}}}.\frac{\beta (\frac{3}{4},\frac{1}{4})}{2}=\frac{1}{4a^{\frac{3}{4}}}.\frac{\mathrm{\Gamma }\left(\frac{1}{4}\right)\mathrm{\Gamma }\left(\frac{3}{4}\right)}{\mathrm{\Gamma }\left(1\right)}=\frac{\pi }{4a^{\frac{3}{4}}sin\left(\frac{\pi }{4}\right)}$$$$=\frac{\pi \sqrt{2}}{4a^{\frac{3}{4}}}=f\left(a\right)$$$$g\left(a\right)=-f^{\prime }\left(a\right)=\frac{3\pi \sqrt{2}}{16a^{\frac{7}{4}}}.$$$${\int }_0^{+\mathrm{\infty }}\frac{dx}{x^4+1}=f\left(1\right)=\frac{\pi \sqrt{2}}{4}=\frac{\pi }{2\sqrt{2}}$$$${\int }_0^{+\mathrm{\infty }}\frac{dx}{2x^4+8}=\frac{f\left(4\right)}{2}=\frac{\pi \sqrt{2}}{8\left(2^{\frac{3}{2}}\right)}=\frac{\pi }{16}$$$$\int \frac{dx}{{(1+x^4)}^2}=g\left(1\right)=-f^{\prime }\left(1\right)=\frac{3\pi \sqrt{2}}{16}$$$${\int }_0^{+\mathrm{\infty }}\frac{dx}{{(2x^4+8)}^2}=\frac{1}{4}g\left(4\right)=-\frac{f^{\prime }\left(4\right)}{4}=\frac{3\pi \sqrt{2}}{64.4^{\frac{7}{4}}}$$$$=\frac{3\pi }{512}$$$$$$

Commented by mathmax by abdo last updated on 19/Mar/20

$$thankyousir.$$

Commented by mind is power last updated on 19/Mar/20

$$withepleasur$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com