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0-t-n-1-t-t-2-dt-




Question Number 160415 by amin96 last updated on 29/Nov/21
∫_0 ^∞ (t^n /(1+t+t^2 ))dt=?
$$\int_{\mathrm{0}} ^{\infty} \frac{\boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}} }{\mathrm{1}+\boldsymbol{\mathrm{t}}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\boldsymbol{\mathrm{dt}}=? \\ $$
Answered by MJS_new last updated on 29/Nov/21
for n∈N  n=0 ⇒ ∫_0 ^∞ (dt/(t^2 +t+1))=((2(√3))/9)π  n≥1 ⇒ ∫_0 ^∞ (dt/(t^2 +t+1))=+∞
$$\mathrm{for}\:{n}\in\mathbb{N} \\ $$$${n}=\mathrm{0}\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{9}}\pi \\ $$$${n}\geqslant\mathrm{1}\:\Rightarrow\:\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}=+\infty \\ $$
Commented by amin96 last updated on 29/Nov/21
∫(t^n /(1+t+t^2 ))dt=?   so when
$$\int\frac{\boldsymbol{\mathrm{t}}^{\boldsymbol{\mathrm{n}}} }{\mathrm{1}+\boldsymbol{\mathrm{t}}+\boldsymbol{\mathrm{t}}^{\mathrm{2}} }\boldsymbol{\mathrm{dt}}=?\: \\ $$so when

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