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Question Number 142429 by Mathspace last updated on 31/May/21
calculate U_n =∫_0 ^∞ ((log^n x)/(1+x^n ))dx find nature of the serie ΣU_n
$${calculate}\:{U}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{{log}^{{n}} {x}}{\mathrm{1}+{x}^{{n}} }{dx} \\ $$$${find}\:{nature}\:{of}\:{the}\:{serie}\:\Sigma{U}_{{n}} \\ $$
Answered by mathmax by abdo last updated on 12/Jun/21
U_n =∫_0 ^∞ ((log^n (x))/(1+x^n ))dx =_(x=t^(1/n) ) ∫_0 ^∞ ((log^n (t^(1/n) ))/(1+t))(1/n)t^((1/n)−1) dt =(1/n^2 )∫_0 ^∞ (t^((1/n)−1) /(1+t))log^n (t)dt let f(a)=∫_0 ^∞ (t^(a−1) /(1+t))dt ⇒ f(a)=∫_0 ^∞ (e^((a−1)logt) /(1+t))dt ⇒f^((n)) (a) =∫_0 ^∞ (∂^n /∂a^n )((e^((a−1)logt) /(1+t)))dt =∫_0 ^∞ ((t^(a−1) log^n (t))/(1+t))dt ⇒f^((n)) ((1/n))=∫_0 ^∞ ((t^((1/n)−1) log^n (t))/(1+t))dt ⇒ U_n =(1/n^2 )f^((n)) ((1/n)) we have f(a)=(π/(sin(πa))) ⇒ f^((n)) (a)=π((1/(sin(πa))))^((n)) =π(((2i)/(e^(iπa) −e^(−iπa) )))^((n)) =2iπ((1/(e^(iπa) −e^(−iπa) )))^((n)) .....be continued....
$$\mathrm{U}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{n}} \left(\mathrm{x}\right)}{\mathrm{1}+\mathrm{x}^{\mathrm{n}} }\mathrm{dx}\:=_{\mathrm{x}=\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} } \:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{log}^{\mathrm{n}} \left(\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}} \right)}{\mathrm{1}+\mathrm{t}}\frac{\mathrm{1}}{\mathrm{n}}\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{log}^{\mathrm{n}} \left(\mathrm{t}\right)\mathrm{dt}\:\:\mathrm{let}\:\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{f}\left(\mathrm{a}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{logt}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\partial^{\mathrm{n}} }{\partial\mathrm{a}^{\mathrm{n}} }\left(\frac{\mathrm{e}^{\left(\mathrm{a}−\mathrm{1}\right)\mathrm{logt}} }{\mathrm{1}+\mathrm{t}}\right)\mathrm{dt} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\mathrm{a}−\mathrm{1}} \mathrm{log}^{\mathrm{n}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow\mathrm{f}^{\left(\mathrm{n}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{n}}−\mathrm{1}} \mathrm{log}^{\mathrm{n}} \left(\mathrm{t}\right)}{\mathrm{1}+\mathrm{t}}\mathrm{dt}\:\Rightarrow \\ $$$$\mathrm{U}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{n}^{\mathrm{2}} }\mathrm{f}^{\left(\mathrm{n}\right)} \left(\frac{\mathrm{1}}{\mathrm{n}}\right)\:\:\mathrm{we}\:\mathrm{have}\:\mathrm{f}\left(\mathrm{a}\right)=\frac{\pi}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\:\Rightarrow \\ $$$$\mathrm{f}^{\left(\mathrm{n}\right)} \left(\mathrm{a}\right)=\pi\left(\frac{\mathrm{1}}{\mathrm{sin}\left(\pi\mathrm{a}\right)}\right)^{\left(\mathrm{n}\right)} \:=\pi\left(\frac{\mathrm{2i}}{\mathrm{e}^{\mathrm{i}\pi\mathrm{a}} −\mathrm{e}^{−\mathrm{i}\pi\mathrm{a}} }\right)^{\left(\mathrm{n}\right)} \\ $$$$=\mathrm{2i}\pi\left(\frac{\mathrm{1}}{\mathrm{e}^{\mathrm{i}\pi\mathrm{a}} −\mathrm{e}^{−\mathrm{i}\pi\mathrm{a}} }\right)^{\left(\mathrm{n}\right)} …..\mathrm{be}\:\mathrm{continued}…. \\ $$

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