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

∫_0 ^∞ ((√t)/(1+t^2 ))dt FAILED TO CALCULATE

$$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$$$ \\ $$

Commented by GalaxyBills last updated on 17/Apr/22

∫_0 ^∞ ((√t)/(1+t^2 ))dt FAILED TO CALCULATE Solution by Awudulai 🇬🇭 let t=u^2 →dt=2udu ∫_0 ^∞ ((2u(√u^2 ))/(1+u^4 ))du→∫_0 ^∞ ((2u^2 )/(1+u^4 ))du let z=u^4 →^(u=z^(1/4) ) dz=4u^3 ⇒du=(dz/(4u^3 )) ∫_0 ^∞ ((2u^2 .u^(−3) )/(4(1+z)))dz→(1/2)∫_0 ^∞ (z^(−(1/4)) /((1+z)))dz using ∫_0 ^∞ (x^(m−1) /((1+x)^(m+n) ))dx→Γ(m)Γ(n) m=(3/4) amd n=(1/4) (1/2)Γ((3/4))Γ((1/4))→(1/2)((π/(sin((π/4)))))=((π(√2))/2)=((2π)/(2(√2)))=(π/( (√2))) Kudos to my Mentors: Prempeh (Euclid)🇬🇭 and to Daniel (Small Einstein)🇳🇬

$$\int_{\mathrm{0}} ^{\infty} \frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{FAILED}\:\mathrm{TO}\:\mathrm{CALCULATE} \\ $$$$\mathfrak{S}{olution}\:\:{by}\:{Awudulai} \\ $$🇬🇭 $${let}\:{t}={u}^{\mathrm{2}} \rightarrow{dt}=\mathrm{2}{udu} \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{u}\sqrt{{u}^{\mathrm{2}} }}{\mathrm{1}+{u}^{\mathrm{4}} }{du}\rightarrow\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{4}} }{du} \\ $$$${let}\:{z}={u}^{\mathrm{4}} \overset{{u}={z}^{\frac{\mathrm{1}}{\mathrm{4}}} } {\rightarrow}{dz}=\mathrm{4}{u}^{\mathrm{3}} \Rightarrow{du}=\frac{{dz}}{\mathrm{4}{u}^{\mathrm{3}} } \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{u}^{\mathrm{2}} .{u}^{−\mathrm{3}} }{\mathrm{4}\left(\mathrm{1}+{z}\right)}{dz}\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{z}^{−\frac{\mathrm{1}}{\mathrm{4}}} }{\left(\mathrm{1}+{z}\right)}{dz} \\ $$$${using}\:\int_{\mathrm{0}} ^{\infty} \frac{{x}^{{m}−\mathrm{1}} }{\left(\mathrm{1}+{x}\right)^{{m}+{n}} }{dx}\rightarrow\Gamma\left({m}\right)\Gamma\left({n}\right) \\ $$$${m}=\frac{\mathrm{3}}{\mathrm{4}}\:\:{amd}\:{n}=\frac{\mathrm{1}}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\right)=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\mathrm{2}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$Kudos to my Mentors: Prempeh (Euclid)🇬🇭 and to Daniel (Small Einstein)🇳🇬

Answered by Mathspace last updated on 17/Apr/22

I=∫_0 ^∞ ((√t)/(1+t^2 ))dt ((√t)=x) ⇒ I=∫_0 ^∞ (x/(1+x^4 ))(2x)dx =2∫_0 ^∞ (x^2 /(1+x^4 ))dx (x=z^(1/4) ) =2∫_0 ^∞ (z^(1/2) /(1+z))(1/4)z^((1/4)−1) dz =(1/2)∫_0 ^∞ (z^((3/4)−1) /(1+z))dz=(1/2)×(π/(sin(((3π)/4)))) =(π/2).(1/(1/( (√2))))=((π(√2))/2)=(π/( (√2)))

$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\sqrt{{t}}}{\mathrm{1}+{t}^{\mathrm{2}} }{dt}\:\:\left(\sqrt{{t}}={x}\right)\:\Rightarrow \\ $$$${I}=\int_{\mathrm{0}} ^{\infty} \:\frac{{x}}{\mathrm{1}+{x}^{\mathrm{4}} }\left(\mathrm{2}{x}\right){dx} \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{4}} }{dx}\:\:\:\:\:\:\left({x}={z}^{\frac{\mathrm{1}}{\mathrm{4}}} \right) \\ $$$$=\mathrm{2}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{z}^{\frac{\mathrm{1}}{\mathrm{2}}} }{\mathrm{1}+{z}}\frac{\mathrm{1}}{\mathrm{4}}{z}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} {dz} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \:\frac{{z}^{\frac{\mathrm{3}}{\mathrm{4}}−\mathrm{1}} }{\mathrm{1}+{z}}{dz}=\frac{\mathrm{1}}{\mathrm{2}}×\frac{\pi}{{sin}\left(\frac{\mathrm{3}\pi}{\mathrm{4}}\right)} \\ $$$$=\frac{\pi}{\mathrm{2}}.\frac{\mathrm{1}}{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}}=\frac{\pi\sqrt{\mathrm{2}}}{\mathrm{2}}=\frac{\pi}{\:\sqrt{\mathrm{2}}} \\ $$

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