Question Number 54367 by maxmathsup by imad last updated on 02/Feb/19
![find ∫_1 ^(+∞) (([t])/t) t^(−p) dt interms of ξ(p) with p>0 .](Q54367.png)
$${find}\:\int_{\mathrm{1}} ^{+\infty} \:\frac{\left[{t}\right]}{{t}}\:{t}^{−{p}} {dt}\:{interms}\:{of}\:\xi\left({p}\right)\:{with}\:{p}>\mathrm{0}\:. \\ $$
Commented bymaxmathsup by imad last updated on 04/Feb/19
![let A_p =∫_1 ^(+∞) (([t])/t) t^(−p) dt ⇒A_p =Σ_(n=1) ^(+∞) ∫_n ^(n+1) (n/t) t^(−p) dt =Σ_(n=1) ^∞ n ∫_n ^(n+1) t^(−p−1) dt =Σ_(n=1) ^∞ n[(1/(−p)) t^(−p) ]_n ^(n+1) =Σ_(n=1) ^∞ (−(n/p)){ (1/((n+1)^p )) −(1/n^p )} =(1/p){Σ_(n=1) ^∞ (1/n^(p−1) ) −Σ_(n=1) ^∞ (n/((n+1)^p ))} =(1/p){ Σ_(n=1) ^∞ (1/n^(p−1) ) −Σ_(n=2) ^∞ ((n−1)/n^p )} =(1/p){ 1+Σ_(n=2) ^∞ (1/n^p )} =((ξ(p))/p) ⇒ A_p =((ξ(p))/p) .](Q54485.png)
$${let}\:{A}_{{p}} =\int_{\mathrm{1}} ^{+\infty} \:\:\:\frac{\left[{t}\right]}{{t}}\:{t}^{−{p}} \:{dt}\:\Rightarrow{A}_{{p}} =\sum_{{n}=\mathrm{1}} ^{+\infty} \:\:\int_{{n}} ^{{n}+\mathrm{1}} \:\:\:\frac{{n}}{{t}}\:{t}^{−{p}} \:{dt} \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\:\int_{{n}} ^{{n}+\mathrm{1}} \:{t}^{−{p}−\mathrm{1}} {dt}\:=\sum_{{n}=\mathrm{1}} ^{\infty} \:{n}\left[\frac{\mathrm{1}}{−{p}}\:{t}^{−{p}} \right]_{{n}} ^{{n}+\mathrm{1}} \\ $$ $$=\sum_{{n}=\mathrm{1}} ^{\infty} \:\left(−\frac{{n}}{{p}}\right)\left\{\:\frac{\mathrm{1}}{\left({n}+\mathrm{1}\right)^{{p}} }\:−\frac{\mathrm{1}}{{n}^{{p}} }\right\}\:=\frac{\mathrm{1}}{{p}}\left\{\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{p}−\mathrm{1}} }\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{n}}{\left({n}+\mathrm{1}\right)^{{p}} }\right\} \\ $$ $$=\frac{\mathrm{1}}{{p}}\left\{\:\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{p}−\mathrm{1}} }\:−\sum_{{n}=\mathrm{2}} ^{\infty} \:\frac{{n}−\mathrm{1}}{{n}^{{p}} }\right\}\:=\frac{\mathrm{1}}{{p}}\left\{\:\mathrm{1}+\sum_{{n}=\mathrm{2}} ^{\infty} \:\:\frac{\mathrm{1}}{{n}^{{p}} }\right\}\:=\frac{\xi\left({p}\right)}{{p}}\:\Rightarrow \\ $$ $${A}_{{p}} =\frac{\xi\left({p}\right)}{{p}}\:\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
![∫_1 ^∞ (([t])/t)t^(−p) dt ∫_1 ^2 1×t^(−p−1) dt+∫_2 ^3 2×t^(−p−1) dt+∫_3 ^4 3×t^(−p−1) dt... 1×((t^(−p) /(−p)))_1 ^2 +2×((t^(−p) /(−p)))_2 ^3 +3×((t^(−p) /(−p)))_3 ^4 +... =((1/(−p)))[((1/2^p )−(1/1^p ))+2×((1/3^p )−(1/2^p ))+3×((1/4^p )−(1/3^p ))+...] =((1/p))((1/1^p )+(1/2^p )+(1/3^p )+(1/4^p )+...∞)](Q54405.png)
$$\int_{\mathrm{1}} ^{\infty} \frac{\left[{t}\right]}{{t}}{t}^{−{p}} {dt} \\ $$ $$\int_{\mathrm{1}} ^{\mathrm{2}} \mathrm{1}×{t}^{−{p}−\mathrm{1}} {dt}+\int_{\mathrm{2}} ^{\mathrm{3}} \mathrm{2}×{t}^{−{p}−\mathrm{1}} {dt}+\int_{\mathrm{3}} ^{\mathrm{4}} \mathrm{3}×{t}^{−{p}−\mathrm{1}} {dt}... \\ $$ $$\mathrm{1}×\left(\frac{{t}^{−{p}} }{−{p}}\right)_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{2}×\left(\frac{{t}^{−{p}} }{−{p}}\right)_{\mathrm{2}} ^{\mathrm{3}} +\mathrm{3}×\left(\frac{{t}^{−{p}} }{−{p}}\right)_{\mathrm{3}} ^{\mathrm{4}} +... \\ $$ $$=\left(\frac{\mathrm{1}}{−{p}}\right)\left[\left(\frac{\mathrm{1}}{\mathrm{2}^{{p}} }−\frac{\mathrm{1}}{\mathrm{1}^{{p}} }\right)+\mathrm{2}×\left(\frac{\mathrm{1}}{\mathrm{3}^{{p}} }−\frac{\mathrm{1}}{\mathrm{2}^{{p}} }\right)+\mathrm{3}×\left(\frac{\mathrm{1}}{\mathrm{4}^{{p}} }−\frac{\mathrm{1}}{\mathrm{3}^{{p}} }\right)+...\right] \\ $$ $$=\left(\frac{\mathrm{1}}{{p}}\right)\left(\frac{\mathrm{1}}{\mathrm{1}^{{p}} }+\frac{\mathrm{1}}{\mathrm{2}^{{p}} }+\frac{\mathrm{1}}{\mathrm{3}^{{p}} }+\frac{\mathrm{1}}{\mathrm{4}^{{p}} }+...\infty\right) \\ $$