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## Homework Statement

The question is to get Fourier sine series of e^-x =f(x) on 0<x<1

## Homework Equations

Bn = 2/L ∫ (e^-x) * sin(nπx/L) over the limits 1 to 0, where L = 1

f(x) = summation of Bn*sin(nπx/L)

## The Attempt at a Solution

So I integrated ∫ by part integration

so I took u = sin(nπx) and v = -e^-x and u' = nπ*cos(nπx)

so equation was

(-e^-x)*sin(nπx) - ∫-(e^-x)*nπ*(cos(nπx))

then again I did part integration where u = nπ*cos(nπx), v = e^-x and u' = -n^2*π^2*sin(nπx)

so I got

(e^-x)*nπ*(cos(nπx)) + ∫(e^-x)*n^2*π^2*sin(nπx)

∫(e^-x)*sin(nπx) = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx)) - ∫(e^-x)*n^2*π^2*sin(nπx)

Now taking ∫(e^-x)*sin(nπx) = L i can write the above equation as

L = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx)) - n^2*π^2 L

So L = (-e^-x)*sin(nπx) - (e^-x)*nπ*(cos(nπx))/(1+n^2*π^2)

Can someone please confirm if my integration is correct? Many thanks. I will only put the limits after someone confirms the integration is correct.