calculate-0-dx-x-4-1-2-

Question Number 126493 by mathmax by abdo last updated on 21/Dec/20

$$\mathrm{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Answered by Dwaipayan Shikari last updated on 21/Dec/20

∫_0 ^∞ (dx/((x^4 +1)^2 )) x^4 =t⇒4x^3 =(dt/dx) =(1/4)∫_0 ^∞ (t^(−(3/4)) /((t+1)^2 ))dt (t/(t+1))=u =(1/4)∫_0 ^1 ((u/(1−u)))^(−(3/4)) du = (1/4)β((1/4),(7/4))=((3Γ((1/4))Γ((3/4)))/(16))=((3π)/(8(√2)))

$$\int_{\mathrm{0}} ^{\infty} \frac{{dx}}{\left({x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\:\:\:\:\:{x}^{\mathrm{4}} ={t}\Rightarrow\mathrm{4}{x}^{\mathrm{3}} =\frac{{dt}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\infty} \frac{{t}^{−\frac{\mathrm{3}}{\mathrm{4}}} }{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{dt}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{t}}{{t}+\mathrm{1}}={u} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{u}}{\mathrm{1}−{u}}\right)^{−\frac{\mathrm{3}}{\mathrm{4}}} {du}\:\:=\:\frac{\mathrm{1}}{\mathrm{4}}\beta\left(\frac{\mathrm{1}}{\mathrm{4}},\frac{\mathrm{7}}{\mathrm{4}}\right)=\frac{\mathrm{3}\Gamma\left(\frac{\mathrm{1}}{\mathrm{4}}\right)\Gamma\left(\frac{\mathrm{3}}{\mathrm{4}}\right)}{\mathrm{16}}=\frac{\mathrm{3}\pi}{\mathrm{8}\sqrt{\mathrm{2}}} \\ $$

Answered by mathmax by abdo last updated on 21/Dec/20

let f(a) =∫_0 ^∞ (dx/(x^4 +a^4 )) with a>0 ⇒f^′ (a)=−∫_0 ^∞ ((4a^3 )/((x^4 +a^4 )^2 ))dx ⇒ f^′ (1)=−4∫_0 ^∞ (dx/((x^4 +1)^2 )) ⇒∫_0 ^∞ (dx/((x^4 +1)^2 ))=−(1/4)f^′ (1) let explicit f(a) f(a) =_(x=at) ∫_0 ^∞ ((adt)/(a^4 (t^4 +1))) =(1/a^3 )∫_0 ^∞ (dt/(1+t^4 )) =_(t=z^(1/4) ) =(1/(4a^3 ))∫_0 ^∞ (1/(1+z))z^((1/4)−1) dz =(1/(4a^3 ))×(π/(sin((π/4)))) =(π/(4a^3 ×((√2)/2)))=(π/(2(√2)a^3 )) ⇒ f(a)=(π/(2(√2)))a^(−3) ⇒f^′ (a) =(π/(2(√2)))(−3)a^(−4) ⇒f^′ (1)=((−3π)/(2(√2))) ⇒ ∫_0 ^∞ (dx/((x^4 +1)^2 )) =−(1/4)×((−3π)/(2(√2))) =((3π)/(8(√2))) =((3π(√2))/(16))

$$\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\mathrm{x}^{\mathrm{4}} +\mathrm{a}^{\mathrm{4}} }\:\:\mathrm{with}\:\mathrm{a}>\mathrm{0}\:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)=−\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{4a}^{\mathrm{3}} }{\left(\mathrm{x}^{\mathrm{4}} \:+\mathrm{a}^{\mathrm{4}} \right)^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow \\ $$$$\mathrm{f}^{'} \left(\mathrm{1}\right)=−\mathrm{4}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }=−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{f}^{'} \left(\mathrm{1}\right)\:\mathrm{let}\:\mathrm{explicit}\:\mathrm{f}\left(\mathrm{a}\right) \\ $$$$\mathrm{f}\left(\mathrm{a}\right)\:=_{\mathrm{x}=\mathrm{at}} \:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{adt}}{\mathrm{a}^{\mathrm{4}} \left(\mathrm{t}^{\mathrm{4}} +\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{dt}}{\mathrm{1}+\mathrm{t}^{\mathrm{4}} }\:\:=_{\mathrm{t}=\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{4}}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{3}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+\mathrm{z}}\mathrm{z}^{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{1}} \mathrm{dz}\:=\frac{\mathrm{1}}{\mathrm{4a}^{\mathrm{3}} }×\frac{\pi}{\mathrm{sin}\left(\frac{\pi}{\mathrm{4}}\right)}\:=\frac{\pi}{\mathrm{4a}^{\mathrm{3}} ×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}\mathrm{a}^{\mathrm{3}} }\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{a}^{−\mathrm{3}} \:\Rightarrow\mathrm{f}^{'} \left(\mathrm{a}\right)\:=\frac{\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\left(−\mathrm{3}\right)\mathrm{a}^{−\mathrm{4}} \:\Rightarrow\mathrm{f}^{'} \left(\mathrm{1}\right)=\frac{−\mathrm{3}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}\right)^{\mathrm{2}} }\:=−\frac{\mathrm{1}}{\mathrm{4}}×\frac{−\mathrm{3}\pi}{\mathrm{2}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{3}\pi}{\mathrm{8}\sqrt{\mathrm{2}}}\:=\frac{\mathrm{3}\pi\sqrt{\mathrm{2}}}{\mathrm{16}} \\ $$

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