Question Number 162238 by amin96 last updated on 27/Dec/21
$$\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{\boldsymbol{\mathrm{x}}^{\mathrm{7}} +\mathrm{1}}\boldsymbol{\mathrm{dx}}=? \\ $$
Answered by Ar Brandon last updated on 27/Dec/21
![z^7 +1=0⇒z_k =e^(((2k+1)/7)iπ) , k∈[0, 6], k∈Z I=∫_0 ^1 (1/(x^7 +1))dx=∫_0 ^1 (dz/(Π_(k=0) ^6 (z−z_k )))=∫_0 ^1 Σ_(k=0) ^6 (a_k /(z−z_k ))dz a_k =(1/(7z_k ^6 ))=−(z_k /7) I=−(1/7)Σ_(k=0) ^6 ∫_0 ^1 (z_k /(z−z_k ))dz=−Σ_(k=0) ^6 [((z_k ln∣z−z_k ∣)/7)]_0 ^1](https://www.tinkutara.com/question/Q162239.png)
$${z}^{\mathrm{7}} +\mathrm{1}=\mathrm{0}\Rightarrow{z}_{{k}} ={e}^{\frac{\mathrm{2}{k}+\mathrm{1}}{\mathrm{7}}{i}\pi} ,\:{k}\in\left[\mathrm{0},\:\mathrm{6}\right],\:{k}\in\mathbb{Z} \\ $$$${I}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{\mathrm{1}}{{x}^{\mathrm{7}} +\mathrm{1}}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dz}}{\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\prod}}\left({z}−{z}_{{k}} \right)}=\int_{\mathrm{0}} ^{\mathrm{1}} \underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\frac{{a}_{{k}} }{{z}−{z}_{{k}} }{dz} \\ $$$${a}_{{k}} =\frac{\mathrm{1}}{\mathrm{7}{z}_{{k}} ^{\mathrm{6}} }=−\frac{{z}_{{k}} }{\mathrm{7}} \\ $$$${I}=−\frac{\mathrm{1}}{\mathrm{7}}\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{z}_{{k}} }{{z}−{z}_{{k}} }{dz}=−\underset{{k}=\mathrm{0}} {\overset{\mathrm{6}} {\sum}}\left[\frac{{z}_{{k}} \mathrm{ln}\mid{z}−{z}_{{k}} \mid}{\mathrm{7}}\right]_{\mathrm{0}} ^{\mathrm{1}} \\ $$