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

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?

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  • $\begingroup$ The standard first year calculus method is to inttegrate by parts twice. $\endgroup$ – André Nicolas Nov 24 '14 at 20:30
  • $\begingroup$ First year calculus? I'm not even in university yet... But thanks, I'll give that a go. $u=sinx$ and $dv=e^-x dx$ right? edit: one of the answers has just checked this, thanks. $\endgroup$ – Jim Nov 24 '14 at 20:35
  • $\begingroup$ By first-year calculus I meant the standard North American course, which is a first exposure to calculus. $\endgroup$ – André Nicolas Nov 24 '14 at 20:38
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$a>0$ \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*} $$\Leftrightarrow I=\frac{b}{a^2+b^2}$$

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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 $I = \frac{\text{some stuff}}{1+c}$.

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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?

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  • $\begingroup$ I haven't learnt about complex exponentials; is there another way to do it? $\endgroup$ – Jim Nov 24 '14 at 20:29
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    $\begingroup$ It would probably be less messy to write $\sin x = \Im e^{ix}$. $\endgroup$ – Najib Idrissi Nov 24 '14 at 20:29
  • $\begingroup$ @Jim: If you don't know complex exponentials, I think integration by parts is the way you're expected to proceed. (André beat me to it :P) $\endgroup$ – Eric Stucky Nov 24 '14 at 20:30
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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.

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$$\int_{0}^{\infty}x^n\exp(-x)dx = n!\Longrightarrow$$

$$\int_{0}^{\infty}\sin(\lambda x)\exp(-x)dx = \sum_{n=0}^{\infty}(-1)^n\lambda^{2n+1}=\frac{\lambda}{1+\lambda^2}$$

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  • $\begingroup$ I feel like this needs more explanation, stating that you are using Taylor series, for how the factorials cancel, and the final sum evaluation, for those unfamiliar with geometric series. It is very clever, though. +1 $\endgroup$ – Bennett Gardiner Nov 26 '14 at 3:29
  • $\begingroup$ Do we need $|\lambda|<1$? $\endgroup$ – Math-fun Nov 26 '14 at 19:14
  • $\begingroup$ Yes, $|\lambda|<1$ in the derivation, and then analytic contination to complex $\lambda$ with magnitide of imaginary part less than 1 will do. The integral exists and is analytic in that region, so, the usual argument that if two analytic functions are equal to each other in a bounded region containing a limit point, then they must be equal to each everywhere in that region will then do. $\endgroup$ – Count Iblis Nov 26 '14 at 19:24
  • $\begingroup$ Many thanks. Let $\lambda=2$, then $\sum_{n=0}^{\infty}(-1)^n2^{2n+1}$ seems not be convergent, or maybe I do not get the point. $\endgroup$ – Math-fun Nov 26 '14 at 19:37
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    $\begingroup$ For $\lambda$ with modulus less than 1 it is convergent, so for these cases the result is valid. The next step is to note that the integral does exist for all real $\lambda$ and is an anlytic function of $\lambda$. So, it is some analytic function that must be equal to $\frac{\lambda}{1+\lambda^2}$ when $|\lambda|<1$. This implies that the integral is in fact equal to $\frac{\lambda}{1+\lambda^2}$ because analytic continuations are unique (you can't have two different analytic functions that are equal to each other in a region containing a limit point). $\endgroup$ – Count Iblis Nov 26 '14 at 20:02
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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!

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  • $\begingroup$ I think you mean product rule/integration by parts rather than chain rule? I feel like your indefinite integral is probably wrong, if I differentiate the RHS w.r.t $x$, I should get the integrand, but I fail to see where the exponentials would come from. The usual way of doing this is to introduce the parameter $\alpha$ in the exponential. $\endgroup$ – Bennett Gardiner Nov 26 '14 at 3:26
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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}

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