an interesting integral

[@maxvangulik1988]

I=sum[k=0,♾️]((-1)^k•int[kpi,(k+1)pi](e^-x•sin(x))dx)
sin(x)=(e^ix-e^-ix)/2i
I=-i/2•sum[k=0,♾️]((-1)^k•int[kpi,(k+1)pi](e^(-1+i)x-e^(-1-i)x)dx)
I=-i/2•sum[k=0,♾️]((-1)^k•((e^(-1+i)(k+1)pi-e^(-1+i)kpi)/(-1+i)-(e^(-1-i)(k+1)pi-e^(-1-i)kpi)/(-1-i)))
e^+-ikpi=(-1)^k
I=-i/2•sum[k=0,♾️](e^-kpi•((e^(-1+i)pi-1)/(-1+i)-(e^(-1-i)pi-1)/(-1-i))
e^ipi=-1
I=sum[k=0,♾️](e^-kpi)•((e^-pi+1)/(1-i)-(e^-pi+1)/(1+i))/2i
(1+i)(1-i)=2
I=sum[k=0,♾️](e^-kpi)•(e^-pi+1)/2
I=(e^-pi+1)/2•1/(1-e^-pi)
I=1/2•(1+e^-pi)/(1-e^-pi)
I=1/2•coth(pi/2)

[@maxvangulik1988]

I=sum[k=0,♾️]((-1)^k•int[kpi,(k+1)pi](e^-x•sin(x))dx)
sin(x)=Im(e^ix)
I=Im(sum[k=0,♾️]((-1)^k•int[kpi,(k+1)pi](e^(-1+i)x)dx))
I=Im(sum[k=0,♾️]((-1)^k•(e^(-1+i)x/(-1+i))|[kpi,(k+1)pi]))
e^inpi=(-1)^n, n€Z
I=Im(sum[k=0,♾️]((-1)^k•(e^-(k+1)pi•(-1)^(k+1)-e^-kpi•(-1)^k)/(-1+i)))
I=Im(sum[k=0,♾️]((e^-(k+1)pi+e^-kpi)/(1-i)))
1/(1-i)=(1+i)/2
I=1/2•sum[k=0,♾️](e^-(k+1)pi+e^-kpi)
I=1/2•(e^-pi/(1-e^-pi)+1/(1-e^-pi))
I=1/2•(1+e^-pi)/(1-e^-pi)
I=1/2•coth(pi/2)
I swear I've solved this one before.

[@maxvangulik1988]

i did. It wasn't showing up for some reason.