Question Number 100653 by Ar Brandon last updated on 28/Jun/20

Answered by mathmax by abdo last updated on 28/Jun/20

1) f(x) =((sin(nx))/(sinx)) is continue  on]0,(π/2)] ⇒ integrable at V(0) we have  f(x)∼((nx)/x) =n so  I_n exist  2)I_n  −I_(n−2) =∫_0 ^(π/2)  ((sin(nx)−sin(n−2)x)/(sinx))dx  sinp −sinq =sinp +sin(−q) =cos((π/2)−p) +cos((π/2) +q)  =2cos(((π−p+q)/2)) cos(((−p−q)/2)) =2cos((π/2)−((p−q)/2))cos(((p+q)/2))  =2cos(((p+q)/2))sin(((p−q)/2)) ⇒sin(nx)−sin(n−2)x =2 cos(((nx+nx−2x)/2))sin(((nx−nx+2x)/2))  =2cos(n−1)x sinx ⇒I_n −I_(n−2) =2∫_0 ^(π/2)  cos(n−1)x dx  =2[(1/(n−1))sin(n−1)x]_0 ^(π/2)  =(2/(n−1))sin(n−1)(π/2) =(2/(n−1)) sin(((nπ)/2)−(π/2))  =−(2/(n−1))cos(((nπ)/2))   (n≥2) ⇒I_2 −I_0 =−(2/1)cos(π) =2 but I_0 =0 ⇒I_2 =2  3)we have I_n −I_(n−2) =−(2/(n−1)) cos(((nπ)/2)) ⇒I_(2n) −I_(2n−2) =−((2(−1)^n )/(n−1)) ⇒  Σ_(k=2) ^n  (I_(2k) −I_(2k−2) ) =2 Σ_(k=2) ^n  (((−1)^(k−1) )/(k−1)) ⇒  I_4  −I_2  +I_6 −I_4  +....I_(2n)  −I_(2n−2) =2 Σ_(k=2) ^n  (((−1)^(k−1) )/(k−1)) ⇒  I_(2n) =I_2  +2Σ_(k=2) ^n  (((−1)^(k−1) )/(k−1)) =2 +2Σ_(k=1) ^(n−1)  (((−1)^k )/k)   I_(2n+1) −I_(2n−1) =−(2/(2n)) cos((((2n+1)π)/2)) =−(1/n)cos(nπ +(π/2)) =0 ⇒  I_(2n+1) =I_(2n−1)  =I_1 =(π/2)

Commented bymathmax by abdo last updated on 28/Jun/20

error at I_(2n)     we have I_n −I_(n−2) =((−2(−1)^n )/(n−1)) ⇒I_(2n) −I_(2n−2) =((−2(−1)^n )/(2n−1))  ⇒Σ_(k=1) ^n  (I_(2k) −I_(2k−2) ) =2 Σ_(k=1) ^n  (((−1)^(k−1) )/(2k−1)) ⇒  I_(2n) −I_0 =2  Σ_(k=1) ^n  (((−1)^(k−1) )/(2k−1)) but I_0 =0 ⇒ I_(2n) =2 Σ_(k=1) ^n  (((−1)^(k−1) )/(2k−1))  =2 Σ_(k=0) ^(n−1)  (((−1)^k )/(2k+1)) and we see thst lim_(n→+∞)  I_(2n) =(π/2)

Commented byAr Brandon last updated on 28/Jun/20

Thank you Sir 😃