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Question Number 25682 by abdo imad last updated on 13/Dec/17

we give ∫_0 ^∞ t^(a−1) (1 + t)^(−1) dt =π (sin(πa))^(−1) with 0<a<1 find the value of ∫_0 ^∞ (1 +x^(16) )^(−1) dx

$${we}\:{give}\:\int_{\mathrm{0}} ^{\infty} \:{t}^{{a}−\mathrm{1}} \left(\mathrm{1}\:+\:{t}\right)^{−\mathrm{1}} {dt}\:=\pi\:\left({sin}\left(\pi{a}\right)\right)^{−\mathrm{1}} \:{with}\:\mathrm{0}<{a}<\mathrm{1}\:{find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\left(\mathrm{1}\:+{x}^{\mathrm{16}} \right)^{−\mathrm{1}} {dx} \\ $$

Answered by ajfour last updated on 13/Dec/17

If x^(16) =t and let ∫(dx/(1+x^(16) )) = I Then I=∫(dt/(16x^(15) (1+t))) =(1/(16))∫_0 ^( ∞) ((t^(−15/16) dt)/(1+t)) =(1/(16))∫_0 ^( ∞) ((t^((1/(16))−1) dt)/(1+t)) =(1/(16))×(π/(sin ((π/(16))))) sin (π/(16)) =(√((1−cos (π/8))/2)) =(√((1−(√((1+1/(√2))/2)))/2)) .

$${If}\:\:{x}^{\mathrm{16}} =\boldsymbol{{t}}\:\:{and}\:\:\:{let}\:\:\int\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{16}} }\:=\:{I} \\ $$ $${Then}\:\:{I}=\int\frac{{dt}}{\mathrm{16}{x}^{\mathrm{15}} \left(\mathrm{1}+{t}\right)} \\ $$ $$\:\:\:=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\:\:\infty} \:\frac{{t}^{−\mathrm{15}/\mathrm{16}} \:{dt}}{\mathrm{1}+{t}}\: \\ $$ $$\:\:=\frac{\mathrm{1}}{\mathrm{16}}\int_{\mathrm{0}} ^{\:\:\infty} \:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{16}}−\mathrm{1}} {dt}}{\mathrm{1}+{t}}\:=\frac{\mathrm{1}}{\mathrm{16}}×\frac{\pi}{\mathrm{sin}\:\left(\frac{\pi}{\mathrm{16}}\right)} \\ $$ $$\mathrm{sin}\:\frac{\pi}{\mathrm{16}}\:=\sqrt{\frac{\mathrm{1}−\mathrm{cos}\:\left(\pi/\mathrm{8}\right)}{\mathrm{2}}}\: \\ $$ $$\:\:\:\:\:\:\:\:\:\:\:\:\:=\sqrt{\frac{\mathrm{1}−\sqrt{\frac{\mathrm{1}+\mathrm{1}/\sqrt{\mathrm{2}}}{\mathrm{2}}}}{\mathrm{2}}}\:\:. \\ $$

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