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Question Number 168743 by mehdiAz last updated on 16/Apr/22

$${\int }_0^{\mathrm{\infty }}\frac{\sqrt{t}}{1+t^2}dt$$$$\mathrm{FAILED}\mathrm{TO}\mathrm{CALCULATE}$$$$$$

Commented by GalaxyBills last updated on 17/Apr/22

$${\int }_0^{\mathrm{\infty }}\frac{\sqrt{t}}{1+t^2}dt$$$$\mathrm{FAILED}\mathrm{TO}\mathrm{CALCULATE}$$$$\mathfrak{S}olutionbyAwudulai$$🇬🇭 $$lett=u^2\to dt=2udu$$$${\int }_0^{\mathrm{\infty }}\frac{2u\sqrt{u^2}}{1+u^4}du\to {\int }_0^{\mathrm{\infty }}\frac{2u^2}{1+u^4}du$$$$letz=u^4\stackrel{u=z^{\frac{1}{4}}}{\to }dz=4u^3\Rightarrow du=\frac{dz}{4u^3}$$$${\int }_0^{\mathrm{\infty }}\frac{2u^2.u^{-3}}{4(1+z)}dz\to \frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{z^{-\frac{1}{4}}}{(1+z)}dz$$$$using{\int }_0^{\mathrm{\infty }}\frac{x^{m-1}}{{(1+x)}^{m+n}}dx\to \mathrm{\Gamma }\left(m\right)\mathrm{\Gamma }\left(n\right)$$$$m=\frac{3}{4}amdn=\frac{1}{4}$$$$\frac{1}{2}\mathrm{\Gamma }\left(\frac{3}{4}\right)\mathrm{\Gamma }\left(\frac{1}{4}\right)\to \frac{1}{2}\left(\frac{\pi }{sin\left(\frac{\pi }{4}\right)}\right)=\frac{\pi \sqrt{2}}{2}=\frac{2\pi }{2\sqrt{2}}=\frac{\pi }{\sqrt{2}}$$Kudos to my Mentors: Prempeh (Euclid)🇬🇭 and to Daniel (Small Einstein)🇳🇬

Answered by Mathspace last updated on 17/Apr/22

$$I={\int }_0^{\mathrm{\infty }}\frac{\sqrt{t}}{1+t^2}dt(\sqrt{t}=x)\Rightarrow$$$$I={\int }_0^{\mathrm{\infty }}\frac{x}{1+x^4}\left(2x\right)dx$$$$=2{\int }_0^{\mathrm{\infty }}\frac{x^2}{1+x^4}dx(x=z^{\frac{1}{4}})$$$$=2{\int }_0^{\mathrm{\infty }}\frac{z^{\frac{1}{2}}}{1+z}\frac{1}{4}{z}^{\frac{1}{4}-1}dz$$$$=\frac{1}{2}{\int }_0^{\mathrm{\infty }}\frac{z^{\frac{3}{4}-1}}{1+z}dz=\frac{1}{2}\times \frac{\pi }{sin\left(\frac{3\pi }{4}\right)}$$$$=\frac{\pi }{2}.\frac{1}{\frac{1}{\sqrt{2}}}=\frac{\pi \sqrt{2}}{2}=\frac{\pi }{\sqrt{2}}$$

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