Question Number 28613 by abdo imad last updated on 27/Jan/18
$${let}\:{give}\:{x}>\mathrm{0}\:\:{and}\:{S}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{sint}}{{e}^{{xt}} −\mathrm{1}}{dt}\:. \\ $$ $${developp}\:{S}\:{at}\:{form}\:{of}\:{series}. \\ $$
Commented byabdo imad last updated on 01/Feb/18
![S(x)= ∫_0 ^∞ ((e^(−xt) sint)/(1−e^(−xt) ))dt= ∫_0 ^∞ ( Σ_(n=0) ^∞ e^(−nxt) )e^(−xt) sintdt = Σ_(n=0) ^∞ ∫_0 ^∞ e^(−(n+1)xt) sint dt=Σ_(n=0) ^∞ A_n (x) A_n (x)= ∫_0 ^∞ e^(−(n+1)xt) sint dt=−Im( ∫_0 ^∞ e^(−((n+1)x +i)t) dt)but ∫_0 ^∞ e^(−((n+1)x+i)t) dt =−(1/((n+1)x+i))[ e^(−((n+1)x+i)t) ]_0 ^(+∞) = (1/((n+1)x+i))=(((n+1)x−i)/((n+1)^2 x^2 +1))=(((n+1)x)/(1+(n+1)^2 x^2 ))−(i/(1+(n+1)^2 x^2 )) A_n (x)= (1/(1+(n+1)^2 x^2 )) ⇒S(x)= Σ_(n=0) ^∞ (1/(1+(n+1)^2 x^2 )) with x>0](Q28944.png)
$${S}\left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:\frac{{e}^{−{xt}} \:{sint}}{\mathrm{1}−{e}^{−{xt}} }{dt}=\:\int_{\mathrm{0}} ^{\infty} \left(\:\sum_{{n}=\mathrm{0}} ^{\infty} {e}^{−{nxt}} \right){e}^{−{xt}} {sintdt} \\ $$ $$=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){xt}} {sint}\:{dt}=\sum_{{n}=\mathrm{0}} ^{\infty} \:{A}_{{n}} \left({x}\right)\: \\ $$ $${A}_{{n}} \left({x}\right)=\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left({n}+\mathrm{1}\right){xt}} {sint}\:{dt}=−{Im}\left(\:\int_{\mathrm{0}} ^{\infty} \:{e}^{−\left(\left({n}+\mathrm{1}\right){x}\:+{i}\right){t}} {dt}\right){but} \\ $$ $$\int_{\mathrm{0}} ^{\infty} \:\:{e}^{−\left(\left({n}+\mathrm{1}\right){x}+{i}\right){t}} {dt}\:=−\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){x}+{i}}\left[\:\:{e}^{−\left(\left({n}+\mathrm{1}\right){x}+{i}\right){t}} \right]_{\mathrm{0}} ^{+\infty} \\ $$ $$=\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right){x}+{i}}=\frac{\left({n}+\mathrm{1}\right){x}−{i}}{\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{1}}=\frac{\left({n}+\mathrm{1}\right){x}}{\mathrm{1}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }−\frac{{i}}{\mathrm{1}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} } \\ $$ $${A}_{{n}} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }\:\Rightarrow{S}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+\left({n}+\mathrm{1}\right)^{\mathrm{2}} {x}^{\mathrm{2}} }\:\:{with}\:{x}>\mathrm{0} \\ $$ $$ \\ $$