Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 64166 by mathmax by abdo last updated on 14/Jul/19

calculate  A_n =∫_(−∞) ^(+∞)   (dx/((x^2 +1)(x^2  +2)....(x^2  +n)))  with n integr natural  and n≥1

$${calculate}\:\:{A}_{{n}} =\int_{−\infty} ^{+\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+\mathrm{2}\right)....\left({x}^{\mathrm{2}} \:+{n}\right)} \\ $$$${with}\:{n}\:{integr}\:{natural}\:\:{and}\:{n}\geqslant\mathrm{1} \\ $$

Answered by Eminem last updated on 14/Jul/19

f(z)=(dz/((z^2 +1)......(z^2 +n)))  A_n =2iπRe(zk=i(√k),k∈[1..n])  res(f.i(√k))=(1/(2i(√k)Π_(j=1) ^(n,k≠j) (1−j)))=((1−k)/(2i(√k)(−1)^n (n−1)!))  A_n =2iπΣ_(k=2) ^n (((1−k))/(2i(√(k ))(n−1)!))+((2iπ)/(2i(−1)^n (n−1)!))

$${f}\left({z}\right)=\frac{{dz}}{\left({z}^{\mathrm{2}} +\mathrm{1}\right)......\left({z}^{\mathrm{2}} +{n}\right)} \\ $$$${A}_{{n}} =\mathrm{2}{i}\pi{Re}\left({zk}={i}\sqrt{{k}},{k}\in\left[\mathrm{1}..{n}\right]\right) \\ $$$${res}\left({f}.{i}\sqrt{{k}}\right)=\frac{\mathrm{1}}{\mathrm{2}{i}\sqrt{{k}}\underset{{j}=\mathrm{1}} {\overset{{n},{k}\neq{j}} {\prod}}\left(\mathrm{1}−{j}\right)}=\frac{\mathrm{1}−{k}}{\mathrm{2}{i}\sqrt{{k}}\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)!} \\ $$$${A}_{{n}} =\mathrm{2}{i}\pi\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}\frac{\left(\mathrm{1}−{k}\right)}{\mathrm{2}{i}\sqrt{{k}\:}\left({n}−\mathrm{1}\right)!}+\frac{\mathrm{2}{i}\pi}{\mathrm{2}{i}\left(−\mathrm{1}\right)^{{n}} \left({n}−\mathrm{1}\right)!} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com