Question Number 226935 by Estevao last updated on 19/Dec/25
Commented by Kassista last updated on 21/Dec/25
Commented by Kassista last updated on 21/Dec/25
Answered by Estevao last updated on 19/Dec/25
$$\boldsymbol{{PROVE}} \\ $$
Answered by Ghisom_ last updated on 20/Dec/25
![I (n)=∫_0 ^∞ (dx/(x^n +1))= [t=(1/(x^n +1)) → dx=−t^(−(1/n)−1) (1−t)^((1/n)−1) (dt/n)] =(1/n)∫_0 ^∞ t^(−(1/n)) (1−t)^((1/n)−1) dt= =(1/n)∫_0 ^∞ t^((1−(1/n))−1) (1−t)^((1/n)−1) dt= =(1/n)B ((1/n), 1−(1/n)) =((Γ ((1/n)) Γ (1−(1/n)))/(nΓ ((1/n)+1−(1/n))))= =((Γ ((1/n)) Γ (1−(1/n)))/n)= =(π/(nsin (π/n)))=I (n) ⇒ ∫_0 ^∞ ((5dx)/(x^(10) +1))=5I (10) = π](https://www.tinkutara.com/question/Q226944.png)
$${I}\:\left({n}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dx}}{{x}^{{n}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\mathrm{1}}{{x}^{{n}} +\mathrm{1}}\:\rightarrow\:{dx}=−{t}^{−\frac{\mathrm{1}}{{n}}−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \frac{{dt}}{{n}}\right] \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{−\frac{\mathrm{1}}{{n}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dt}= \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} {dt}= \\ $$$$=\frac{\mathrm{1}}{{n}}\mathrm{B}\:\left(\frac{\mathrm{1}}{{n}},\:\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\:=\frac{\Gamma\:\left(\frac{\mathrm{1}}{{n}}\right)\:\Gamma\:\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)}{{n}\Gamma\:\left(\frac{\mathrm{1}}{{n}}+\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)}= \\ $$$$=\frac{\Gamma\:\left(\frac{\mathrm{1}}{{n}}\right)\:\Gamma\:\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)}{{n}}= \\ $$$$=\frac{\pi}{{n}\mathrm{sin}\:\frac{\pi}{{n}}}={I}\:\left({n}\right) \\ $$$$\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{5}{dx}}{{x}^{\mathrm{10}} +\mathrm{1}}=\mathrm{5}{I}\:\left(\mathrm{10}\right)\:=\:\pi \\ $$
Commented by Ghisom_ last updated on 20/Dec/25
![J (m, n)=∫_0 ^∞ (x^m /(x^n +1))dx= [same substitution as above] =(1/n)∫_0 ^∞ t^((1−(m/n)−(1/n))−1) (1−t)^(((m/n)+(1/n))−1) dt= =(1/n)B (((m+1)/n), 1−((m+1)/n)) = =(π/(nsin ((π(m+1))/n)))=J (m, n) ⇒ ∫_0 ^∞ ((5x^2 )/(x^(10) +1))dx=5J (2, 10) =(π/ )](https://www.tinkutara.com/question/Q226945.png)
$${J}\:\left({m},\:{n}\right)=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{x}^{{m}} }{{x}^{{n}} +\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[\mathrm{same}\:\mathrm{substitution}\:\mathrm{as}\:\mathrm{above}\right] \\ $$$$=\frac{\mathrm{1}}{{n}}\underset{\mathrm{0}} {\overset{\infty} {\int}}{t}^{\left(\mathrm{1}−\frac{{m}}{{n}}−\frac{\mathrm{1}}{{n}}\right)−\mathrm{1}} \left(\mathrm{1}−{t}\right)^{\left(\frac{{m}}{{n}}+\frac{\mathrm{1}}{{n}}\right)−\mathrm{1}} {dt}= \\ $$$$=\frac{\mathrm{1}}{{n}}\mathrm{B}\:\left(\frac{{m}+\mathrm{1}}{{n}},\:\mathrm{1}−\frac{{m}+\mathrm{1}}{{n}}\right)\:= \\ $$$$=\frac{\pi}{{n}\mathrm{sin}\:\frac{\pi\left({m}+\mathrm{1}\right)}{{n}}}={J}\:\left({m},\:{n}\right) \\ $$$$\Rightarrow \\ $$$$\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{5}{x}^{\mathrm{2}} }{{x}^{\mathrm{10}} +\mathrm{1}}{dx}=\mathrm{5}{J}\:\left(\mathrm{2},\:\mathrm{10}\right)\:=\frac{\pi}{\:} \\ $$