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0-1-x-49-1-x-x-2-x-3-x-100-dx-




Question Number 155619 by mathdanisur last updated on 02/Oct/21
𝛀 =∫_( 0) ^( 1)  (x^(49) /(1 + x + x^2  + x^3  ... x^(100) )) dx = ?
$$\boldsymbol{\Omega}\:=\underset{\:\mathrm{0}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\mathrm{x}^{\mathrm{49}} }{\mathrm{1}\:+\:\mathrm{x}\:+\:\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{x}^{\mathrm{3}} \:…\:\mathrm{x}^{\mathrm{100}} }\:\mathrm{dx}\:=\:? \\ $$
Answered by mindispower last updated on 04/Oct/21
=∫_0 ^1 ((x^(49) (1−x))/(1−x^(50) ))dx  u=x^(50) ⇒x^(49) dx=(du/(50))  =∫_0 ^1 (((1−u^(1/(50)) ))/(1−u))(du/(50))  =(1/(50))∫_0 ^1 ((1−u^(1/(50)) )/(1−u))du  Ψ(1+x)=−γ+∫_0 ^1 ((1−t^x )/(1−t))dx  Ω=(1/(50))(Ψ(((51)/(50)))+γ).
$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{49}} \left(\mathrm{1}−{x}\right)}{\mathrm{1}−{x}^{\mathrm{50}} }{dx} \\ $$$${u}={x}^{\mathrm{50}} \Rightarrow{x}^{\mathrm{49}} {dx}=\frac{{du}}{\mathrm{50}} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{1}−{u}^{\frac{\mathrm{1}}{\mathrm{50}}} \right)}{\mathrm{1}−{u}}\frac{{du}}{\mathrm{50}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{50}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{u}^{\frac{\mathrm{1}}{\mathrm{50}}} }{\mathrm{1}−{u}}{du} \\ $$$$\Psi\left(\mathrm{1}+{x}\right)=−\gamma+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}−{t}^{{x}} }{\mathrm{1}−{t}}{dx} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{50}}\left(\Psi\left(\frac{\mathrm{51}}{\mathrm{50}}\right)+\gamma\right). \\ $$
Commented by puissant last updated on 04/Oct/21
∫(x^(49) /(1+x+...+x^(100) ))dx=∫((x^(49) (1−x))/((1−x^(101) )))dx≠∫((x^(49) (1−x))/((1−x^(50) )))dx
$$\int\frac{{x}^{\mathrm{49}} }{\mathrm{1}+{x}+…+{x}^{\mathrm{100}} }{dx}=\int\frac{{x}^{\mathrm{49}} \left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}^{\mathrm{101}} \right)}{dx}\neq\int\frac{{x}^{\mathrm{49}} \left(\mathrm{1}−{x}\right)}{\left(\mathrm{1}−{x}^{\mathrm{50}} \right)}{dx} \\ $$
Commented by mindispower last updated on 04/Oct/21
changed quation it was 1+x...+x^(49)
$${changed}\:{quation}\:{it}\:{was}\:\mathrm{1}+{x}…+{x}^{\mathrm{49}} \\ $$
Commented by puissant last updated on 04/Oct/21
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