Question and Answers Forum

All Questions      Topic List

Relation and Functions Questions

Previous in All Question      Next in All Question      

Previous in Relation and Functions      Next in Relation and Functions      

Question Number 130136 by mathmax by abdo last updated on 22/Jan/21

calculate  ∫_0 ^∞   ((x^2 lnx)/(1+x^6 ))dx

$$\mathrm{calculate}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx} \\ $$

Answered by Dwaipayan Shikari last updated on 22/Jan/21

I′(2)=((−π^2 )/6^2 ).((cos(((2+1)/6)π))/(sin^2 (((2+1)/6)π)))=0 (Proved Earlier)

$${I}'\left(\mathrm{2}\right)=\frac{−\pi^{\mathrm{2}} }{\mathrm{6}^{\mathrm{2}} }.\frac{{cos}\left(\frac{\mathrm{2}+\mathrm{1}}{\mathrm{6}}\pi\right)}{{sin}^{\mathrm{2}} \left(\frac{\mathrm{2}+\mathrm{1}}{\mathrm{6}}\pi\right)}=\mathrm{0}\:\left({Proved}\:{Earlier}\right) \\ $$

Answered by Ar Brandon last updated on 22/Jan/21

I=∫_0 ^1 ((x^2 lnx)/(1+x^6 ))dx+∫_1 ^∞ ((x^2 lnx)/(1+x^6 ))dx     =∫_0 ^1 ((x^2 lnx)/(1+x^6 ))dx−∫_0 ^1 ((t^2 lnt)/(1+t^2 ))dt=0

$$\mathcal{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx}+\int_{\mathrm{1}} ^{\infty} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx} \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx}−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{t}^{\mathrm{2}} \mathrm{lnt}}{\mathrm{1}+\mathrm{t}^{\mathrm{2}} }\mathrm{dt}=\mathrm{0} \\ $$

Answered by mathmax by abdo last updated on 23/Jan/21

I=∫_0 ^∞  ((x^2 lnx)/(1+x^6 ))dx  we do the changement x^6  =t ⇒  I =(1/6)∫_0 ^∞   (((t^(1/6) )^2  ln(t^(1/6) ))/(1+t)) t^((1/6)−1)  dt =(1/(36))∫_0 ^∞  ((t^((1/3)+(1/6)−1) ln(t))/(1+t))dt  =(1/(36))∫_0 ^∞  ((t^(−(1/2)) ln(t))/(1+t))dt  but  ∫_0 ^∞   ((t^(a−1) lnt)/(1+t))dt =−((π^2 cos(πa))/(sin^2 (πa))) ⇒  ∫_0 ^∞ ((t^(−(1/2)) ln(t))/(1+t))dt =−((π^2 cos((π/2)))/(sin^2 ((π/2))))=0 ⇒I=0

$$\mathrm{I}=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{x}^{\mathrm{2}} \mathrm{lnx}}{\mathrm{1}+\mathrm{x}^{\mathrm{6}} }\mathrm{dx}\:\:\mathrm{we}\:\mathrm{do}\:\mathrm{the}\:\mathrm{changement}\:\mathrm{x}^{\mathrm{6}} \:=\mathrm{t}\:\Rightarrow \\ $$$$\mathrm{I}\:=\frac{\mathrm{1}}{\mathrm{6}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{\left(\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)^{\mathrm{2}} \:\mathrm{ln}\left(\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{6}}} \right)}{\mathrm{1}+\mathrm{t}}\:\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \:\mathrm{dt}\:=\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{6}}−\mathrm{1}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{36}}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\:\mathrm{but}\:\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{lnt}}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\pi\mathrm{a}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{ln}\left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:=−\frac{\pi^{\mathrm{2}} \mathrm{cos}\left(\frac{\pi}{\mathrm{2}}\right)}{\mathrm{sin}^{\mathrm{2}} \left(\frac{\pi}{\mathrm{2}}\right)}=\mathrm{0}\:\Rightarrow\mathrm{I}=\mathrm{0} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com