Vector Questions

Question Number 100850 by Dwaipayan Shikari last updated on 28/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{102}} \left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right).....\left({x}−\mathrm{100}\right)×\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}+\frac{\mathrm{1}}{{x}−\mathrm{2}}+...+\frac{\mathrm{1}}{{x}−\mathrm{100}}\right){dx} \\$$

Commented byDwaipayan Shikari last updated on 28/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{102}} \frac{{d}}{{dx}}\left(\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right).....\left({x}−\mathrm{100}\right)\right){dx} \\$$ $$\int_{\mathrm{0}} ^{\mathrm{102}} {d}\left(\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right).....\left({x}−\mathrm{100}\right)\right) \\$$ $$\left[\left({x}−\mathrm{1}\right)\left({x}−\mathrm{2}\right)....\left({x}−\mathrm{100}\right)\underset{\mathrm{0}} {\overset{\mathrm{102}} {\right]}} \\$$ $$\:\:\:\:=\mathrm{101}!−\mathrm{100}! \\$$ $$\\$$

Answered by maths mind last updated on 28/Jun/20

$$\int_{\mathrm{0}} ^{\mathrm{102}} \frac{\Gamma\left({x}\right)}{\Gamma\left({x}−\mathrm{100}\right)}\left(\Psi\left({x}\right)−\Psi\left({x}−\mathrm{100}\right)\right){dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{102}} \frac{\Gamma\left({x}\right)}{\Gamma\left({x}−\mathrm{100}\right)}\Psi\left({x}\right)−\int_{\mathrm{0}} ^{\mathrm{102}} \frac{\Gamma\left({x}\right)\Psi\left({x}−\mathrm{100}\right)}{\Gamma\left({x}−\mathrm{100}\right)} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{102}} \frac{\Gamma'\left({x}\right)\Gamma\left({x}−\mathrm{100}\right)−\Gamma'\left({x}−\mathrm{100}\right)\Gamma\left({x}\right)}{\Gamma\left({x}−\mathrm{100}\right)^{\mathrm{2}} }{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{102}} {d}\left(\frac{\Gamma\left({x}\right)}{\Gamma\left({x}−\mathrm{100}\right)}\right) \\$$ $$=\left[\frac{\Gamma\left({x}\right)}{\Gamma\left({x}−\mathrm{100}\right)}\right]_{\mathrm{0}} ^{\mathrm{102}} =\frac{\Gamma\left(\mathrm{102}\right)}{\Gamma\left(\mathrm{2}\right)}−\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\Gamma\left({a}\right)}{\Gamma\left({a}−\mathrm{100}\right)} \\$$ $$\Gamma\left({a}\right)=\left({a}−\mathrm{1}\right)......\left({a}−\mathrm{100}\right)\Gamma\left({a}−\mathrm{100}\right) \\$$ $$\Rightarrow\frac{\Gamma\left({a}\right)}{\Gamma\left({a}−\mathrm{100}\right)}=\left(\mathrm{100}−{a}\right)....\left(\mathrm{1}−{a}\right) \\$$ $$\Rightarrow\underset{{a}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\Gamma\left({a}\right)}{\Gamma\left({a}−\mathrm{100}\right)}=\Gamma\left(\mathrm{101}\right) \\$$ $${we}\:{get}\:\Gamma\left(\mathrm{102}\right)−\Gamma\left(\mathrm{101}\right)=\mathrm{100}\Gamma\left(\mathrm{101}\right) \\$$ $$\\$$ $$\\$$ $$\\$$ $$\\$$ $$\\$$

Commented bymaths mind last updated on 28/Jun/20

$${we}\:{can}\:{get}\:\int_{\mathrm{0}} ^{{k}+\mathrm{2}} \underset{{j}=\mathrm{0}} {\overset{{k}} {\prod}}\left({x}−{j}\right).\left(\underset{{j}=\mathrm{0}} {\overset{{k}} {\sum}}\frac{\mathrm{1}}{{x}−{j}}\right){dx} \\$$ $$={k}\Gamma\left({k}+\mathrm{1}\right) \\$$

Answered by mathmax by abdo last updated on 28/Jun/20

$$\mathrm{let}\:\mathrm{p}\left(\mathrm{x}\right)\:=\prod_{\mathrm{k}=\mathrm{1}} ^{\mathrm{100}} \left(\mathrm{x}−\mathrm{k}\right)\:\Rightarrow\frac{\mathrm{p}^{'} \left(\mathrm{x}\right)}{\mathrm{p}\left(\mathrm{x}\right)}\:=\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{100}} \:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{k}}\:\Rightarrow \\$$ $$\mathrm{p}\left(\mathrm{x}\right)\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{100}} \:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{k}}\:=\mathrm{p}^{'} \left(\mathrm{x}\right)\:\Rightarrow\int_{\mathrm{0}} ^{\mathrm{102}} \left(\mathrm{x}−\mathrm{1}\right)\left(\mathrm{x}−\mathrm{2}\right)...\left(\mathrm{x}−\mathrm{100}\right)×\left(\frac{\mathrm{1}}{\mathrm{x}−\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{x}−\mathrm{2}}+...+\frac{\mathrm{1}}{\mathrm{x}−\mathrm{100}}\right)\mathrm{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{102}} \mathrm{p}^{'} \left(\mathrm{x}\right)\mathrm{dx}\:=\left[\mathrm{p}\left(\mathrm{x}\right)\right]_{\mathrm{0}} ^{\mathrm{102}} \:=\mathrm{p}\left(\mathrm{102}\right)−\mathrm{p}\left(\mathrm{0}\right) \\$$ $$=\mathrm{101}×\mathrm{100}×...×\mathrm{2}−\mathrm{1}×\mathrm{2}×\mathrm{3}×...\mathrm{100}\:=\left(\mathrm{101}\right)!−\mathrm{100}! \\$$