In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \
We solve the given integral by integration by parts method
u = {e^{ - x}}\ and dv = \sin \left( {nx} \right)\,dx\
du = - {e^{ - x}}dx\ and v=−ncos(nx)
Then
In = \int\limits_0^\infty {{e^{ - x}}\sin \left( {nx} \right)\,dx} \
In=[−ne−xcos(nx)]0∞−0∫∞(−ncos(nx))(−e−x)dx
In=−n1[excos(nx)]0∞−n10∫∞e−xcos(nx)dx
In=−n1[e∞cos(n(∞))−e(0)cos(n(0))]−n10∫∞e−xcos(nx)dx
In=−n1[0−1]−n10∫∞e−xcos(nx)dx
In=n1−n10∫∞e−xcos(nx)dx
\begin{array}{l}
In = \frac{1}{n} - \frac{1}{n}\int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \\
\end{array}\
\begin{array}{l}
In = \frac{1}{n} - \frac{1}{n}\left( I \right) & & & ... & \left( 1 \right)
\end{array}\
Now consider
I = \int\limits_0^\infty {{e^{ - x}}\cos \left( {nx} \right)\,dx} \
Solving by integration by parts
u=e−x and dv=cos(nx)dx
du = - {e^{ - x}}dx\ and v=nsin(nx)
Then
I=[ne−xsin(nx)]0∞−0∫∞n−e−xsin(nx)dx
I=n1[exsin(nx)]0∞+n10∫∞e−xsin(nx)dx
I=n1[0−0]0∞+n1(In)
I=n1(In)
Hence from (1), we have
In=n1−n1(n1(In))
In=n1−n21(In)
In(1+n21)=n1
In=n1×1+n2n2
In=1+n2n
Therefore,
0∫∞e−xsin(nx)dx=1+n2n
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