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I-t-s-0-x-t-1-x-s-dx-1-t-s-t-1-s-I-t-s-0-x-t-1-x-s-dx-0-x-t-1-x-m-s-1-x-m-dx-0-x-t-e-m-s-ln-1-x-1-x-

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Question Number 138378 by Ñï= last updated on 12/Apr/21
I(t,s)=∫_0 ^∞ x^(−t) (1+x)^(−s) dx=((Γ(1−t)Γ(s+t−1))/(Γ(s))) I(t,s)=∫_0 ^∞ x^(−t) (1+x)^(−s) dx =∫_0 ^∞ x^(−t) (1+x)^(m−s) (1+x)^(−m) dx =∫_0 ^∞ x^(−t) e^((m−s)ln(1+x)) (1+x)^(−m) dx =Σ_(n=0) ^∞ (((m−s)^n )/(n!))∫_0 ^∞ ((ln^n (1+x))/(x^t (1+x)^m ))dx =2Σ_(n=0) ^∞ (((m−s)^n )/(n!))∫_0 ^∞ ((ln^n (1+x^2 ))/((1+x^2 )^m ))dx (t=(1/2)) If i want to calculate ∫_0 ^∞ ((ln^n (1+x^2 ))/((1+x^2 )^m )) how to expand “ ((Γ(1−t)Γ(s+t−1))/(Γ(s))) ”into summation?
$${I}\left({t},{s}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{−{t}} \left(\mathrm{1}+{x}\right)^{−{s}} {dx}=\frac{\Gamma\left(\mathrm{1}−{t}\right)\Gamma\left({s}+{t}−\mathrm{1}\right)}{\Gamma\left({s}\right)} \\ $$$${I}\left({t},{s}\right)=\int_{\mathrm{0}} ^{\infty} {x}^{−{t}} \left(\mathrm{1}+{x}\right)^{−{s}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\infty} {x}^{−{t}} \left(\mathrm{1}+{x}\right)^{{m}−{s}} \left(\mathrm{1}+{x}\right)^{−{m}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\int_{\mathrm{0}} ^{\infty} {x}^{−{t}} {e}^{\left({m}−{s}\right){ln}\left(\mathrm{1}+{x}\right)} \left(\mathrm{1}+{x}\right)^{−{m}} {dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({m}−{s}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{{n}} \left(\mathrm{1}+{x}\right)}{{x}^{{t}} \left(\mathrm{1}+{x}\right)^{{m}} }{dx} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left({m}−{s}\right)^{{n}} }{{n}!}\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{{n}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{m}} }{dx}\:\:\:\:\:\:\:\:\:\:\:\:\left({t}=\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$${If}\:{i}\:{want}\:{to}\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \frac{{ln}^{{n}} \left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{m}} } \\ $$$${how}\:{to}\:{expand}\:\:“\:\frac{\Gamma\left(\mathrm{1}−{t}\right)\Gamma\left({s}+{t}−\mathrm{1}\right)}{\Gamma\left({s}\right)}\:''{into}\:{summation}? \\ $$
Answered by Dwaipayan Shikari last updated on 13/Apr/21
∫_0 ^∞ (x^(−t) /((x+1)^s ))dx =∫_0 ^∞ (x^(1−t−1) /((x+1)^(s−1+1−t+t) ))dx =B(s+t−1,1−t) =∫_0 ^1 x^(−t) (1−x)^(s+t−2) dx =Σ_(n=1) ^∞ ∫_0 ^1 (((2−t−s)_n )/(n!))x^(n−t) dx =Σ_(n=1) ^∞ (((2−t−s)_n )/(n!(n−t+1))) =(1/(1−t))Σ_(n=1) ^∞ (((2−t−s)_n (1−t)_n )/((2−t)_n n!)) =(1/(1−t)) _2 F_1 (2−t−s,1−t;2−t;1)
$$\int_{\mathrm{0}} ^{\infty} \frac{{x}^{−{t}} }{\left({x}+\mathrm{1}\right)^{{s}} }{dx} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \frac{{x}^{\mathrm{1}−{t}−\mathrm{1}} }{\left({x}+\mathrm{1}\right)^{{s}−\mathrm{1}+\mathrm{1}−{t}+{t}} }{dx} \\ $$$$={B}\left({s}+{t}−\mathrm{1},\mathrm{1}−{t}\right) \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {x}^{−{t}} \left(\mathrm{1}−{x}\right)^{{s}+{t}−\mathrm{2}} {dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\left(\mathrm{2}−{t}−{s}\right)_{{n}} }{{n}!}{x}^{{n}−{t}} \:{dx} \\ $$$$=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}−{t}−{s}\right)_{{n}} }{{n}!\left({n}−{t}+\mathrm{1}\right)}\:=\frac{\mathrm{1}}{\mathrm{1}−{t}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}−{t}−{s}\right)_{{n}} \left(\mathrm{1}−{t}\right)_{{n}} }{\left(\mathrm{2}−{t}\right)_{{n}} {n}!} \\ $$$$ \\ $$$$=\frac{\mathrm{1}}{\mathrm{1}−{t}}\:_{\mathrm{2}} {F}_{\mathrm{1}} \left(\mathrm{2}−{t}−{s},\mathrm{1}−{t};\mathrm{2}−{t};\mathrm{1}\right) \\ $$

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