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Question-200619

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Question Number 200619 by sonukgindia last updated on 21/Nov/23
Answered by witcher3 last updated on 21/Nov/23
I_9 =2∫_0 ^∞ (x^a /(c+kx^b ))dx in ∞ c+kx^b ∼x^b ;(x^a /(c+kx^b ))∼x^(a−b) integrabl ⇒a−b<−1 ⇒b>1+a ⇒b≥a+2 (k/c)x^b =t⇒x=(((ct)/k))^(1/b) ⇒dx=((c/k))^(1/b) (t^((1/b)−1) /b) I_9 =2∫_0 ^∞ ((t^((1/b)−1) .(((ct)/k))^(a/b) .((c/k))^(1/b) )/(c(1+t)b)) =(2/(cb)).((c/k))^((a+1)/b) ∫_0 ^∞ (t^(((1+a)/b)−1) /(1+t))dt =(2/(cb))((c/k))^(((a+1)/b) ) .β(((1+a)/b),1−((1+a)/b)) =(2/(cb))((c/k))^((a+1)/b) .(π/(sin(((1+a)/b)π)))
$$\mathrm{I}_{\mathrm{9}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{c}+\mathrm{kx}^{\mathrm{b}} }\mathrm{dx} \\ $$$$\mathrm{in}\:\infty\:\mathrm{c}+\mathrm{kx}^{\mathrm{b}} \sim\mathrm{x}^{\mathrm{b}} ;\frac{\mathrm{x}^{\mathrm{a}} }{\mathrm{c}+\mathrm{kx}^{\mathrm{b}} }\sim\mathrm{x}^{\mathrm{a}−\mathrm{b}} \:\mathrm{integrabl}\:\Rightarrow\mathrm{a}−\mathrm{b}<−\mathrm{1} \\ $$$$\Rightarrow\mathrm{b}>\mathrm{1}+\mathrm{a}\:\Rightarrow\mathrm{b}\geqslant\mathrm{a}+\mathrm{2} \\ $$$$\frac{\mathrm{k}}{\mathrm{c}}\mathrm{x}^{\mathrm{b}} =\mathrm{t}\Rightarrow\mathrm{x}=\left(\frac{\mathrm{ct}}{\mathrm{k}}\right)^{\frac{\mathrm{1}}{\mathrm{b}}} \Rightarrow\mathrm{dx}=\left(\frac{\mathrm{c}}{\mathrm{k}}\right)^{\frac{\mathrm{1}}{\mathrm{b}}} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{b}}−\mathrm{1}} }{\mathrm{b}} \\ $$$$\mathrm{I}_{\mathrm{9}} =\mathrm{2}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}}{\mathrm{b}}−\mathrm{1}} .\left(\frac{\mathrm{ct}}{\mathrm{k}}\right)^{\frac{\mathrm{a}}{\mathrm{b}}} .\left(\frac{\mathrm{c}}{\mathrm{k}}\right)^{\frac{\mathrm{1}}{\mathrm{b}}} }{\mathrm{c}\left(\mathrm{1}+\mathrm{t}\right)\mathrm{b}} \\ $$$$=\frac{\mathrm{2}}{\mathrm{cb}}.\left(\frac{\mathrm{c}}{\mathrm{k}}\right)^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}}} \int_{\mathrm{0}} ^{\infty} \frac{\mathrm{t}^{\frac{\mathrm{1}+\mathrm{a}}{\mathrm{b}}−\mathrm{1}} }{\mathrm{1}+\mathrm{t}}\mathrm{dt} \\ $$$$=\frac{\mathrm{2}}{\mathrm{cb}}\left(\frac{\mathrm{c}}{\mathrm{k}}\right)^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}}\:} .\beta\left(\frac{\mathrm{1}+\mathrm{a}}{\mathrm{b}},\mathrm{1}−\frac{\mathrm{1}+\mathrm{a}}{\mathrm{b}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{cb}}\left(\frac{\mathrm{c}}{\mathrm{k}}\right)^{\frac{\mathrm{a}+\mathrm{1}}{\mathrm{b}}} .\frac{\pi}{\mathrm{sin}\left(\frac{\mathrm{1}+\mathrm{a}}{\mathrm{b}}\pi\right)} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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