Menu Close

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

Tinku Tara 0 Comments
Pin




Question Number 133038 by liberty last updated on 18/Feb/21
∫_0 ^1 ((x^4 (1−x)^4 )/(1+x^2 )) dx =?
$$\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:\frac{{x}^{\mathrm{4}} \left(\mathrm{1}−{x}\right)^{\mathrm{4}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:=? \\ $$
Commented by Dwaipayan Shikari last updated on 18/Feb/21
((22)/7)−π
$$\frac{\mathrm{22}}{\mathrm{7}}−\pi \\ $$
Answered by mathmax by abdo last updated on 20/Feb/21
I=∫_0 ^1 ((x^4 (1−x)^4 )/(1+x^2 ))dx ⇒I =∫_0 ^1 x^4 (1−x)^4 Σ_(n=0) ^∞ (−1)^n x^(2n) dx =Σ_(n=0) ^∞ (−1)^n ∫_0 ^1 x^(2n+4) (1−x)^4 dx =Σ_(n=0) ^∞ (−1)^n B(2n+5,5) =Σ_(n=0) ^∞ (−1)^n .((Γ(2n+5).Γ(5))/(Γ(2n+10))) =4! Σ_(n=0) ^∞ (((−1)^n (2n+4)!)/((2n+9)!)) =4! Σ_(n=0) ^∞ (((−1)^n )/((2n+9)(2n+8)(2n+7)(2n+6)(2n+5))) rest to find the value of this serie be continued....
$$\mathrm{I}=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{x}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} }{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\mathrm{dx}\:\Rightarrow\mathrm{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{x}^{\mathrm{4}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} \sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{x}^{\mathrm{2n}} \mathrm{dx} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:\mathrm{x}^{\mathrm{2n}+\mathrm{4}} \left(\mathrm{1}−\mathrm{x}\right)^{\mathrm{4}} \:\mathrm{dx}\:=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{B}\left(\mathrm{2n}+\mathrm{5},\mathrm{5}\right) \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:.\frac{\Gamma\left(\mathrm{2n}+\mathrm{5}\right).\Gamma\left(\mathrm{5}\right)}{\Gamma\left(\mathrm{2n}+\mathrm{10}\right)}\:=\mathrm{4}!\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \left(\mathrm{2n}+\mathrm{4}\right)!}{\left(\mathrm{2n}+\mathrm{9}\right)!} \\ $$$$=\mathrm{4}!\:\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2n}+\mathrm{9}\right)\left(\mathrm{2n}+\mathrm{8}\right)\left(\mathrm{2n}+\mathrm{7}\right)\left(\mathrm{2n}+\mathrm{6}\right)\left(\mathrm{2n}+\mathrm{5}\right)} \\ $$$$\mathrm{rest}\:\mathrm{to}\:\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{this}\:\mathrm{serie}\:\mathrm{be}\:\mathrm{continued}…. \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *