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Question Number 217761 by mnjuly1970 last updated on 20/Mar/25
prove that : I=∫_0 ^( ∞) ((sin(πx)sin(2πx)sin(3πx))/x^3 ) = π^3
provethat:I=∫0∞sin(πx)sin(2πx)sin(3πx)x3=π3
Answered by Ghisom last updated on 22/Mar/25
∫_0 ^∞ ((sin πx sin 2πx sin 3πx)/x^3 )dx= =(1/4)∫_0 ^∞ (((sin 2πx)/x^3 )+((sin 4πx)/x^3 )−((sin 6πx)/x^3 ))dx J(n)=(1/4)∫_0 ^∞ ((sin nπx)/x^3 )dx= [t=nπx → dx=(dt/(nπ))] =((n^2 π^2 )/4) ∫_0 ^∞ ((sin t)/t^3 )dt= [by parts] =((n^2 π^2 )/8)(−[((sin t)/t^2 )]_0 ^∞ +∫_0 ^∞ ((cos t)/t^2 )dt)= [by parts] =−((n^2 π^2 )/8)([((tcos t +sin t)/t^2 )]_0 ^∞ +∫_0 ^∞ ((sin t)/t)dt)= [lim_(t→0^+ ) ((tcos t +sin t)/t^2 ) =lim_(t→∞) ((tcos t +sin t)/t^2 ) =0] [we know that ∫_0 ^∞ ((sin t)/t)dt=(π/2)] =−((n^2 π^2 )/8)(0+(π/2))=−((n^2 π^3 )/(16)) ⇒ I=J(2)+J(4)−J(6)=−(π^3 /4)−π^3 +((9π^3 )/4)=π^3
∫∞0sinπxsin2πxsin3πxx3dx==14∫∞0(sin2πxx3+sin4πxx3−sin6πxx3)dxJ(n)=14∫∞0sinnπxx3dx=[t=nπx→dx=dtnπ]=n2π24∫∞0sintt3dt=[byparts]=n2π28(−[sintt2]0∞+∫∞0costt2dt)=[byparts]=−n2π28([tcost+sintt2]0∞+∫∞0sinttdt)=[limt→0+tcost+sintt2=limt→∞tcost+sintt2=0][weknowthat∫∞0sinttdt=π2]=−n2π28(0+π2)=−n2π316⇒I=J(2)+J(4)−J(6)=−π34−π3+9π34=π3
Commented by mnjuly1970 last updated on 22/Mar/25
thanks alot sir .very nice solution
thanksalotsir.verynicesolution

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