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Question Number 124228 by mnjuly1970 last updated on 01/Dec/20

$$...:::nicecalculus:::...$$$$evaluate$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{cos\left(ln\left(x\right)\right)}{1+x^3}dx=....???$$

Answered by Lordose last updated on 01/Dec/20

$$$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\frac{\cos \left(ln\left(\mathrm{x}\right)\right)}{1+{\mathrm{x}}^3}\mathrm{dx}$$$$\mathrm{I}={\int }_0^{\mathrm{\infty }}\cos \left(\ln \left(\mathrm{x}\right)\right)\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\mathrm{x}}^{3\mathrm{n}}\mathrm{dx}$$$$\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{3\mathrm{n}}\cos \left(\ln \left(\mathrm{x}\right)\right)\mathrm{dx}$$$$\mathrm{let}-\mathrm{u}=\ln \left(\mathrm{x}\right)\Rightarrow \mathrm{dx}=-\mathrm{xdu}$$$$\mathrm{I}=\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{\mathrm{e}}^{-3\mathrm{nu}-\mathrm{u}}\cos \left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{I}=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}{(-1)}^{\mathrm{n}}{\int }_0^{\mathrm{\infty }}{\mathrm{e}}^{-\mathrm{u}(3\mathrm{n}+1)}\cos \left(\mathrm{u}\right)\mathrm{du}$$$$\mathrm{I}=2\underset{\mathrm{n}=0}{\sum ^{\mathrm{\infty }}}\frac{{(-1)}^{\mathrm{n}}(3\mathrm{n}+1)}{{9\mathrm{n}}^2+6\mathrm{n}+2}$$$$\mathrm{I}=\frac{1}{6}(-{\psi }^0(\frac{1}{6}-\frac{\mathrm{i}}{6})-{\psi }^0(\frac{1}{6}+\frac{\mathrm{i}}{6})+{\psi }^0(\frac{2}{3}-\frac{\mathrm{i}}{6})+{\psi }^0(\frac{2}{3}+\frac{\mathrm{i}}{6}))$$

Answered by mnjuly1970 last updated on 02/Dec/20

$$solution:$$$$\mathrm{I}=Re({\int }_0^{\mathrm{\infty }}\frac{e^{iln\left(x\right)}}{1+x^3}dx=\mathrm{\Omega })$$$$\mathrm{\Omega }={\int }_0^{\mathrm{\infty }}\frac{x^i}{1+x^3}dx\stackrel{(x^3=t)}{=}=\frac{1}{3}{\int }_0^{\mathrm{\infty }}\frac{t^{\frac{i-2}{3}}}{1+t}dt$$$$=\frac{1}{3}\mathrm{\Gamma }\left(\frac{i+1}{3}\right)\mathrm{\Gamma }(1-\frac{1+i}{3})$$$$=\frac{1}{3}\ast \frac{\pi }{sin\left(\frac{(i+1)\pi }{3}\right)}$$$$=\frac{1}{3}\ast \frac{\pi }{\frac{\sqrt{3}}{2}cos\left(\frac{\pi i}{3}\right)+\frac{1}{2}sin\left(\frac{\pi i}{3}\right)}$$$$=\frac{2}{3}\left[\frac{\pi }{\sqrt{3}\left(\frac{e^{-\frac{\pi }{3}}+e^{\frac{\pi }{3}}}{2}\right)-i\left(\frac{e^{\frac{-\pi }{3}}-e^{\frac{\pi }{3}}}{2}\right)}\right]$$$$=\frac{4}{3}\left(\frac{\pi }{\sqrt{3}cos\left(h\left(\frac{\pi }{3}\right)\right)+isin\left(h\left(\frac{\pi }{3}\right)\right)}\right)$$$$=\frac{4\pi }{3}\left\{\frac{\sqrt{3}cos\left(h\left(\frac{\pi }{3}\right)\right)-isin\left(h\left(\frac{\pi }{3}\right)\right)}{3cos\left(h\left(\frac{\pi }{3}\right)\right)+sin\left(h\left(\frac{\pi }{3}\right)\right)}\right\}$$$$\mathrm{I}=Re\left(\mathrm{\Omega }\right)=\frac{4\pi \sqrt{3}}{3}\left(\frac{cos\left(h\left(\frac{\pi }{3}\right)\right)}{3cos\left(h\left(\frac{\pi }{3}\right)\right)+sin\left(h\left(\frac{\pi }{3}\right)\right)}\right)$$$$=\frac{4\pi \sqrt{3}}{3}\left(\frac{1}{3+tan\left(h\left(\frac{\pi }{3}\right)\right)}\right)✓$$$$$$

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