Show that $\int_0^\infty \frac{\sin (\lambda x)}{e^x} \, \mathrm dx =\frac{\lambda}{1+{\lambda^2}}$ 
$$\int_0^\infty \frac{\sin (\lambda x)}{e^x} \, \mathrm dx =\frac{\lambda}{1+{\lambda^2}}$$

My intuition telling me there might be an $\arctan$ coming up, but I don't know how to do this question. Where do I begin?
 A: Assuming you can't use complex exponentials, I would use integration by parts. You will have to do it twice. For the first application of integration by parts, take $u = \sin x$ and $dv = e^{-x} dx$.
Note that, after both applications of integration by parts, if this is done correctly, you should get an equation of the form $I = \text{ some stuff } - cI$, where $I$ is the integral you are trying to evaluate, and $c$ is some constant. Then you can rearrange to get $I + cI = \text{ some stuff }$, or $I = \frac{\text{some stuff}}{1+c}$.
A: $a>0$
\begin{align*}
I&=\int_0^{+\infty}e^{-ax}\sin bx\mathrm{d}x=\int_0^{+\infty}e^{-ax}\mathrm{d}\left(-\frac{\cos bx}{b}\right)\\
&=e^{-ax}\left(-\frac{\cos bx}{b}\right)\mid_0^{+\infty}-\frac{a}{b}\int_0^{+\infty}e^{-ax}\cos bx\mathrm{d}x\\
&=\frac{1}{b}-\frac{a}{b}\int_0^{+\infty}e^{-ax}\mathrm{d}\left(\frac{\sin bx}{b}\right)\\
&=\frac{1}{b}-\frac{a}{b^2}\left(\frac{e^{-ax}\sin bx}{b}\right)\mid_0^{+\infty}-\frac{a^2}{b^2}\int_0^{+\infty}e^{-ax}\sin bc\mathrm{d}x\\
&=\frac{1}{b}-\frac{a^2}{b^2}I
\end{align*}
$$\Leftrightarrow I=\frac{b}{a^2+b^2}$$
A: There are probably other ways to do this, but using complex exponentials is just too nice for me to pass up:
$$\int_0^\infty \dfrac{\sin (\lambda x)}{e^x}dx=\frac1{2i}\int_0^\infty e^{-x+i\lambda x}-e^{-x-i\lambda x} ~dx$$
Can you proceed from here?
A: Let $x_n$, $n=1,2,...$ be such that $\lambda+\tan(x_n\lambda)=0$, with $x_{n+1}>x_n>0$. Let $x_0=0$. Now note that $$\int_0^{\infty}\sin(\lambda x)e^{-x}dx=\sum_{n=0}^{\infty}\int_{x_n}^{x_{n+1}}\sin(\lambda x)e^{-x}dx$$
Using integration-by-part twice we obtain $$\int_0^{x_{1}}\sin(\lambda x)e^{-x}dx=\frac{\lambda-\lambda\cos(\lambda x_1)-\sin(\lambda x_1)}{1+\lambda^2}$$ and that $$\int_{x_n}^{x_{n+1}}\sin(\lambda x)e^{-x}dx=\frac{e^{-x_n}(\lambda \cos(\lambda x_n)+\sin(\lambda x_n))-e^{-x_{n+1}}(\lambda \cos(\lambda x_{n+1})+\sin(\lambda x_{n+1}))}{1+\lambda^2}$$
These complete the proof.
A: 
This is specially for Differentiation-under-Integral-sign lovers

Consider following integral
$$I(\alpha )=\int_0^\infty \frac{\sin (\alpha x)}{e^{x}} \,\mathrm dx=\int_0^\infty e^{-x}\sin (\alpha x)  \,\mathrm dx$$
We have $I(0)=0$
By differentiating under Integral sign wrt $\alpha$
$$I'(\alpha )=\int_0^\infty xe^{-x}\cos (\alpha x) \,\mathrm  dx=\int_0^\infty xe^{x}\cos (\alpha x) \,\mathrm  dx$$

Now consider following Indefinite integral
$$I=\int xe^{x}\cos (\alpha x) \,\mathrm  dx$$
Using Product rule we have
$$\int uv \,\mathrm dw=uvw-\int vw \,\mathrm du-\int uw \,\mathrm dv$$
and then by making following substitutions and proceeding yields$$\begin{align}
x=u&\iff  \,\mathrm dx= \,\mathrm du\\
\cos (\alpha x)=v&\iff -\alpha \sin x \,\mathrm dx= \,\mathrm dv\\
e^x  \,\mathrm dx= \,\mathrm dw&\iff e^x  = w\\
\end{align}$$
$$I=\int xe^{x}\cos (\alpha x)  \,\mathrm dx=e^x\left[\frac{\alpha \left[(\alpha ^2+1)x-2\right]\sin(\alpha x)+\left[\alpha ^2(x+1)+x-1\right]\cos(\alpha x)}{(1+\alpha ^2)^2}\right]$$

$$I'(\alpha )=\int_0^\infty e^{-x}\cos (\alpha x)  \,\mathrm  dx=\frac{1-\alpha ^2}{(1+\alpha ^2)^2}$$
Integrating wrt $\alpha$
$$\begin{align}
I(\alpha )&=\frac{\alpha }{\alpha ^2+1}+c\\
I(0)&=0+c\implies c=0\\
I(\alpha )&=\frac{\alpha }{\alpha ^2+1}\\
\end{align}$$
And Finally!

$$\large I(\alpha )=\int_0^\infty \frac{\sin (\alpha x)}{e^{x}} \,\mathrm  dx=\frac{\alpha }{\alpha ^2+1}$$


I feel so sorry to say that no $\arctan$ is involved here!
A: $$\int_{0}^{\infty}x^n\exp(-x)dx = n!\Longrightarrow$$
$$\int_{0}^{\infty}\sin(\lambda x)\exp(-x)dx = \sum_{n=0}^{\infty}(-1)^n\lambda^{2n+1}=\frac{\lambda}{1+\lambda^2}$$
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}&\color{#66f}{\large%
\int_{0}^{\infty}{\sin\pars{\lambda x} \over \expo{x}}\,\dd x}
=\Im\int_{0}^\infty\expo{\pars{-1 + \ic\lambda}x}\,\dd x
=\Im\pars{1 \over 1 - \ic\lambda}
=\Im\pars{1 + \ic\lambda \over \phantom{\ic\,}1 + \lambda^{2}}
=\color{#66f}{\large{\lambda \over 1 + \lambda^{2}}}
\end{align}
