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I know one can solve this in many ways and the answer is $\pi/4$. However I'm interested in one particular solution involving Laplace transform.

I once saw a solution of this integral where one just did something like directly taking the Laplace transform of the integrand and then getting an integral in terms of $s$. So my question is, how does one go from

$$\mathscr{L}\left[\int_{0}^{\infty}\frac{\sin{x}}{xe^x} \ dx\right]$$

To some integral like

$$\frac{1}{2}\int_0^{\infty}\frac{1}{s^2+1} \ ds = \frac{\pi}{4}.$$

I don't remember the steps inbetween or if the first step is even correct but I wan't to know how the Laplace theory was used here. Any one who has a guess?

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  • $\begingroup$ Maybe this result is helpful (Laplace transform of $\frac{f(t)}{t}$ is $\int_s^\infty F(u)\, du$, where $F=\mathscr{L}(f)$). Your original integral is related to the Laplace transform of $\frac{f(t)}{t}$ where $f(t)=\sin{t}$. $\endgroup$ – Minus One-Twelfth Mar 21 at 22:15
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    $\begingroup$ By the way, I don't think you'd be taking the Laplace transform of your integral, because that would just be the Laplace transform of a constant. $\endgroup$ – Minus One-Twelfth Mar 21 at 22:21
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Something like this? \begin{multline} \int_0^\infty \frac{\sin x}{xe^x} dx = \int_0^\infty \sin x \int_1^\infty e^{-sx}dsdx = \int_1^\infty\!\!\int_0^\infty\sin(x) e^{-sx}dx\,ds \\= \int_1^\infty \mathcal{L}\left[\sin(x)\right](s)ds = \int_1^\infty \frac{ds}{s^2 + 1} = \frac{\pi}{2}-\tan^{-1}1 = \frac{\pi}{4} \end{multline}

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  • $\begingroup$ Your first step, how did you split the integral like that? $\endgroup$ – Parseval Mar 21 at 23:19
  • $\begingroup$ @Parseval Since $ x^{-1}e^{-x}= \int_1^\infty e^{-sx}ds$, one can be freely substituted for the other, by the substitution property of equality. $\endgroup$ – eyeballfrog Mar 22 at 1:33
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Let $$I(a)=\int_0^\infty \frac{\sin(x)}{x}e^{-ax}dx$$ where your integral is $I(1)$. $I'(a)$ is then: $$I'(a)=-\int_0^\infty \sin(x)e^{-ax}dx$$ which is the Laplace Transform of the sine function. Thus $$I'(a)=-\frac{1}{a^2+1}$$

Integrating back we get: $$I(a)=-\arctan(a)+C$$ To find the constant, we notice that $$I(0)=\int_0^\infty \frac{\sin(x)}{x}dx=\frac{\pi}{2}$$ So we get that $C=\frac{\pi}{2}$ and that $$I(1)=-\arctan(1)+\frac{\pi}{2}=\frac{\pi}{4}$$

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Here is an approach that employs the following useful property for the Laplace transform: $$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$ Perhaps it is something like this you are looking for?

Noting that $$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$ and $$\mathcal{L}^{-1} \left \{\frac{e^{-x}}{x} \right \} (t)= H(t - 1).$$ Here $H(x)$ denotes the Heaviside step function. Thus

\begin{align} \int_0^\infty \frac{\sin x}{x e^x} \, dx &= \int_0^\infty \sin x \cdot \frac{e^{-x}}{x} \, dx\\ &= \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{e^{-x}}{x} \right \} (t) \, dt\\ &= \int_0^\infty \frac{H(t - 1)}{1 + t^2} \, dt.\\ &= \int_1^\infty \frac{1}{1 + t^2} \, dt\\ &= \Big{[}\tan^{-1} t \Big{]}_1^\infty\\[1ex] &= \frac{\pi}{2} - \frac{\pi}{4}\\[1ex] &= \frac{\pi}{4}. \end{align}

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The great thing about this one is even "alternative" approaches accidentally at some point get us a double integral whose further evaluation can be seen as using a result in Laplace transforms. For example, aleden's answer begins exactly the way you would if you said "I'd rather use this", but look what it gets!

My usual preferred way of solving this sees the same phenomenon. It begins with a Schwinger parametrization, writing $\frac{1}{x}=\int_0^\infty\exp -sx ds$. Then the integral becomes $\int_0^\infty ds\int_0^\infty dx\sin x\exp -(1+s)x$. Up to a constant variable shift, this yields the deliberately Laplace-based approach in eyeballfrog's answer.

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