Question

نشان دهید که: $$ (-1)^{n+1} G_n= \int _0^ \infty \frac{dt}{ (t+1)^{n} . ( \pi ^{2} + ln^{2} t) } \wedge G_n= \int _0^1 \binom{x}{n}dx $$

Comments

$$G_n= \int _0^1 \frac{ \Gamma (x+1)}{n! \Gamma (x-n+1)}dx \wedge  \Gamma (x)= \frac{ \pi }{sin \pi x} . \frac{1}{ \Gamma (1-x)}   \wedge x \mapsto (x-n+1) \Longrightarrow  \Gamma (x-n+1)= \frac{ \pi }{sin (\pi x).cos \pi (1-n)+cos (\pi x).sin \pi (1-n)} . \frac{1}{ \Gamma (n-x)} = \frac{ \pi }{ \Gamma (n-x).sin \pi x. (-1)^{n+1} } \Longrightarrow  \frac{1}{ \Gamma (x-n+1)} = \frac{ (-1)^{n+1} .sin \pi x. \Gamma (n-x)}{ \pi } \Longrightarrow G_n= (-1)^{n+1}  \int _0^1 \frac{ \Gamma (x+1)}{n!} . \frac{sin \pi x}{ \pi } . \Gamma (n-x)dx= (-1)^{n+1} \int _0^1 \frac{sin \pi x}{ \pi } dx \int _0^1 t^{n-x-1} . (1-t)^{x}  dt= \frac{ (-1)^{n+1} }{ \pi }  \int _0^1 t^{n-1} dt \int _0^1sin( \pi x). ( \frac{1-t}{t} )^{x} dx \wedge m= \frac{1-t}{t}   \Longrightarrow G_n= \frac{ (-1)^{n+1} }{ \pi }  \int _0^1 t^{n-1} dt \int _0^1sin( \pi x). m^{x} $$$$I= \int _0^1sin (\pi x)d( \frac{ m^{x} }{1+m} )= \frac{ m^{x} }{lnm}.sin  \pi x | _0^1- \ \frac{ m^{x} }{lnm}  \int _0^1 m^{x} cos \pi xdx\Longrightarrow I_1=  \int _0^1cos \pi xd (\frac{ m^{x} }{1+m} )= \frac{ m^{x} }{lnm} cos  \pi x | _0^1+ \frac{ \pi }{lnm}  \int _0^1 m^{x} .sin  (\pi x)dx=( \frac{-m}{lnm} - \frac{1}{lnm} )+ \frac{ \pi }{lnm} .I \Longrightarrow I= \frac{ ln^{2} m}{  \pi ^{2} + ln^{2} m} . \frac{ \pi }{ ln^{2}m } (m+1)$$$$replace "m "by  \frac{1-t}{t}  \Longrightarrow I= \frac{m \frac{1}{t} }{  \pi ^{2} . ln^{2}m }  \Longrightarrow G_n= \frac{ (-1)^{n+1} }{ \pi }  \int _0^1 \frac{ t^{1-n} . \pi . \frac{1}{t} }{  \pi ^{2} + ln^{2}( \frac{1}{t} -1) } dt \wedge u= (\frac{1}{t} -1 )\Longrightarrow G_n= (-1)^{n+1}  \int _ { \infty } ^0 \frac{ (u+1)^{-n} }{  \pi ^{2} + ln^{2} u} (-du)= \int _0^ { \infty }  \frac{dt}{ (t+1)^{n} (  \pi ^{2} + ln^{2} t)}  \vee  (-1)^{n+1} G_n= \int _0^ { \infty }  \frac{dt}{ (t+1)^{n} (  \pi ^{2} + ln^{2} t)} $$

Answer 1

$$G_n= \int _0^1 \frac{ \Gamma (x+1)}{n! \Gamma (x-n+1)}dx \wedge \Gamma (x)= \frac{ \pi }{sin \pi x} . \frac{1}{ \Gamma (1-x)} \wedge x \mapsto (x-n+1) \Longrightarrow \Gamma (x-n+1)= \frac{ \pi }{sin (\pi x).cos \pi (1-n)+cos (\pi x).sin \pi (1-n)} . \frac{1}{ \Gamma (n-x)} = \frac{ \pi }{ \Gamma (n-x).sin \pi x. (-1)^{n+1} } \Longrightarrow \frac{1}{ \Gamma (x-n+1)} = \frac{ (-1)^{n+1} .sin \pi x. \Gamma (n-x)}{ \pi } \Longrightarrow G_n= (-1)^{n+1} \int _0^1 \frac{ \Gamma (x+1)}{n!} . \frac{sin \pi x}{ \pi } . \Gamma (n-x)dx= (-1)^{n+1} \int _0^1 \frac{sin \pi x}{ \pi } dx \int _0^1 t^{n-x-1} . (1-t)^{x} dt= \frac{ (-1)^{n+1} }{ \pi } \int _0^1 t^{n-1} dt \int _0^1sin( \pi x). ( \frac{1-t}{t} )^{x} dx \wedge m= \frac{1-t}{t} \Longrightarrow G_n= \frac{ (-1)^{n+1} }{ \pi } \int _0^1 t^{n-1} dt \int _0^1sin( \pi x). m^{x} $$$$I= \int _0^1sin (\pi x)d( \frac{ m^{x} }{1+m} )= \frac{ m^{x} }{lnm}.sin \pi x | _0^1- \ \frac{ m^{x} }{lnm} \int _0^1 m^{x} cos \pi xdx\Longrightarrow I_1= \int _0^1cos \pi xd (\frac{ m^{x} }{1+m} )= \frac{ m^{x} }{lnm} cos \pi x | _0^1+ \frac{ \pi }{lnm} \int _0^1 m^{x} .sin (\pi x)dx=( \frac{-m}{lnm} - \frac{1}{lnm} )+ \frac{ \pi }{lnm} .I \Longrightarrow I= \frac{ ln^{2} m}{ \pi ^{2} + ln^{2} m} . \frac{ \pi }{ ln^{2}m } (m+1)$$$$replace "m "by \frac{1-t}{t} \Longrightarrow I= \frac{m \frac{1}{t} }{ \pi ^{2} . ln^{2}m } \Longrightarrow G_n= \frac{ (-1)^{n+1} }{ \pi } \int _0^1 \frac{ t^{1-n} . \pi . \frac{1}{t} }{ \pi ^{2} + ln^{2}( \frac{1}{t} -1) } dt \wedge u= (\frac{1}{t} -1 )\Longrightarrow G_n= (-1)^{n+1} \int _ { \infty } ^0 \frac{ (u+1)^{-n} }{ \pi ^{2} + ln^{2} u} (-du)= \int _0^ { \infty } \frac{dt}{ (t+1)^{n} ( \pi ^{2} + ln^{2} t)} \vee (-1)^{n+1} G_n= \int _0^ { \infty } \frac{dt}{ (t+1)^{n} ( \pi ^{2} + ln^{2} t)} $$