let-give-0-pi-prove-that-0-1-dt-e-i-t-n-1-e-in-n-

Question Number 28611 by abdo imad last updated on 27/Jan/18

let give θ∈]0,π[ prove that ∫_0 ^1 (dt/(e^(−iθ) −t))= Σ_(n=1) ^(+∞) (e^(inθ) /n) .

$$\left.{let}\:{give}\:\theta\in\right]\mathrm{0},\pi\left[\:\:{prove}\:{that}\:\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{dt}}{{e}^{−{i}\theta} −{t}}=\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\frac{{e}^{{in}\theta} }{{n}}\:\:.\right. \\ $$

Commented by abdo imad last updated on 28/Jan/18

let put I= ∫_0 ^1 (dt/(e^(−iθ) −t)) ⇒ I= ∫_0 ^1 (e^(iθ) /(1−e^(iθ) t)) but we have ∣e^(iθ) t∣≤1 I= ∫_0 ^1 e^(iθ) ( Σ_(n=0) ^(+∞) e^(inθ) t^n )dt = Σ_(n=0) ^(+∞) e^(i(n+1)θ) ∫_0 ^1 t^n dt = Σ_(n=0) ^(+∞) (e^(i(n+1)θ) /(n+1))= Σ_(n=1) ^(+∞) (e^(inθ) /n) .

$${let}\:{put}\:\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{dt}}{{e}^{−{i}\theta} −{t}}\:\Rightarrow\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\frac{{e}^{{i}\theta} }{\mathrm{1}−{e}^{{i}\theta} {t}}\:{but}\:{we}\:{have} \\ $$$$\mid{e}^{{i}\theta} {t}\mid\leqslant\mathrm{1}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{e}^{{i}\theta} \:\left(\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:{e}^{{in}\theta} \:{t}^{{n}} \right){dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:{e}^{{i}\left({n}+\mathrm{1}\right)\theta} \:\int_{\mathrm{0}} ^{\mathrm{1}} \:{t}^{{n}} {dt} \\ $$$$=\:\sum_{{n}=\mathrm{0}} ^{+\infty} \:\:\frac{{e}^{{i}\left({n}+\mathrm{1}\right)\theta} }{{n}+\mathrm{1}}=\:\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{e}^{{in}\theta} }{{n}}\:. \\ $$

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