La solución de esta integral?

Question

podría alguien resolver esta integral ?

$$\int_0^\infty \frac{e^{-x}\sin(x)\cos(ax)}x~\mathrm dx$$

bien, he intentado abrir el pecado*cos utilizando identidades trigonométricas pero eso no ayuda mucho

Answer 1

Tenga en cuenta que podemos escribir

$$\begin{align} e^{-x}\sin(x)\cos(ax)&=\text{Re}\left(e^{-x}e^{iax}\sin(x)\right)\\\\ &=\text{Re}\left(\frac{e^{i(a+1+i)x}-e^{i(a-1+i)x}}{2i}\right) \end{align}$$

Por lo tanto, tenemos desde el Generalizado Frullani del Teorema de

$$\begin{align} \int_0^\infty \frac{e^{-x}\sin(x)\cos(ax)}{x}\,dx&=\text{Re}\left(\frac{1}{2i}\int_0^\infty \frac{e^{i(a+1+i)x}-e^{i(a-1+i)x}}{x}\,dx\right)\\\\ &=\frac12\arctan\left(\frac{\text{Im}((a+1-i)(a-1+i))}{\text{Re}((a+1-i)(a-1+i))}\right)\\\\ &=\frac12 \arctan(2/a^2) \end{align}$$

Answer 2

Por el seno, además de fórmulas, es suficiente para calcular $$f(m)=\int_{0}^{+\infty}\frac{\sin(mx)}{x}e^{-x}\,dx \tag{1}$$ donde $f(0)=0$ y por el teorema de convergencia dominada $$f'(m) = \int_{0}^{+\infty}\cos(mx)e^{-x}\,dx\stackrel{IBP}{=}\frac{1}{1+m^2}\tag{2}$$ lo que implica: $$f(m)=\int_{0}^{+\infty}\frac{\sin(mx)}{x}e^{-x}\,dx = \arctan(m)\tag{3}$$ y $$\begin{eqnarray*} \int_{0}^{+\infty}\frac{\sin(x)\cos(ax)}{x}e^{-x}\,dx &=& \frac{\arctan(1-a)+\arctan(1+a)}{2}\\&=&\color{red}{\frac{1}{2}\,\arctan\frac{2}{a^2}}.\tag{4} \end{eqnarray*}$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove armada]{\displaystyle{#1}}\,} \newcommand{\llaves}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\parcial #3^{#1}}} \newcommand{\raíz}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $$\begin{align} &\int_{0}^{\infty}{\expo{-x}\sin\pars{x}\cos\pars{ax} \over x}\,\dd x \\[5mm] = &\ {1 \over 2} \int_{0}^{\infty}{\expo{-x}\sin\pars{\bracks{1 + a}x} \over x}\,\dd x + {1 \over 2} \int_{0}^{\infty}{\expo{-x}\sin\pars{\bracks{1 - a}x} \over x}\,\dd x \\[5mm] = &\ {1 \over 2}\pars{1 + a} \int_{0}^{\infty}\expo{-x}\, {\sin\pars{\bracks{1 + a}x} \over \pars{1 + a}x}\,\dd x\ +\ \pars{a \leftrightarrow -a} \\[5mm] = &\ {1 \over 2}\pars{1 + a} \int_{0}^{\infty}\expo{-x}\, {1 \over 2}\int_{-1}^{1}\expo{\ic\pars{1 + a}xk}\,\dd k\,\dd x\ +\ \pars{a \leftrightarrow -a} \\[5mm] = &\ {1 \over 4}\pars{1 + a} \int_{-1}^{1}\int_{0}^{\infty}\exp\pars{\bracks{- 1 + \ic\pars{1 + a}k}x} \,\dd x\,\dd k\ +\ \pars{a \leftrightarrow -a} \\[5mm] = &\ {1 \over 4}\pars{1 + a} \int_{-1}^{1}{-1 \over -1 + \ic\pars{1 + a}k}\,\dd k\ +\ \pars{a \leftrightarrow -a} \\[5mm] = &\ {1 \over 4}\pars{1 + a}2 \int_{0}^{1}{\dd k \over \pars{1 + a}^{2}k^{2} + 1}\ +\ \pars{a \leftrightarrow -a} = {1 \over 2} \int_{0}^{1 + a}{\dd k \over k^{2} + 1}\ +\ \pars{a \leftrightarrow -a} \\[5mm] = &\ \bbx{\arctan\pars{1 + a} + \arctan\pars{1 - a} \over 2} \end{align}$$

Answer 4

Así, podemos observar en la transformada de Laplace:

$$\mathscr{I}_{\space\text{a}}\left(\text{s}\right):=\mathscr{L}_x\left[\frac{\sin\left(x\right)\cdot\cos\left(\text{a}\cdot x\right)}{x}\right]_{\left(\text{s}\right)}:=\int_0^\infty\frac{\sin\left(x\right)\cdot\cos\left(\text{a}\cdot x\right)}{x}\cdot e^{-\text{s}\cdot x}\space\text{d}t\tag1$$

Podemos utilizar el 'dominio de la frecuencia de la integración' de la propiedad de la transformada de Laplace:

$$\mathscr{I}_{\space\text{a}}\left(\text{s}\right)=\int_\text{s}^\infty\mathscr{L}_x\left[\sin\left(x\right)\cdot\cos\left(\text{a}\cdot x\right)\right]_{\left(\sigma\right)}\space\text{d}\sigma\tag2$$

Ahora, usted puede utilizar:

$$\sin\left(x\right)\cdot\cos\left(\text{a}\cdot x\right)=\frac{\sin\left(x\cdot\left(1-\text{a}\right)\right)+\sin\left(x\cdot\left(1+\text{a}\right)\right)}{2}\tag3$$

Así, obtenemos:

$$\mathscr{L}_x\left[\sin\left(x\right)\cdot\cos\left(\text{a}\cdot x\right)\right]_{\left(\sigma\right)}=\frac{1}{2}\cdot\left(\mathscr{L}_x\left[\sin\left(x\cdot\left(1-\text{a}\right)\right)\right]_{\left(\sigma\right)}+\mathscr{L}_x\left[\sin\left(x\cdot\left(1+\text{a}\right)\right)\right]_{\left(\sigma\right)}\right)\tag4$$

Ahora, utilice:

• $$\mathscr{L}_x\left[\sin\left(x\cdot\left(1-\text{a}\right)\right)\right]_{\left(\sigma\right)}=\frac{1-\text{a}}{\left(\text{a}-1\right)^2+\sigma^2}\tag5$$
• $$\mathscr{L}_x\left[\sin\left(x\cdot\left(1+\text{a}\right)\right)\right]_{\left(\sigma\right)}=\frac{1+\text{a}}{\left(1+\text{a}\right)^2+\sigma^2}\tag6$$

Así que, finalmente, tenemos que encontrar (resolver el problema):

$$\mathscr{I}_{\space\text{a}}\left(1\right)=\int_1^\infty\frac{1-\text{a}}{\left(\text{a}-1\right)^2+\sigma^2}\cdot\frac{1+\text{a}}{\left(1+\text{a}\right)^2+\sigma^2}\space\text{d}\sigma=$$ $$\left(1-\text{a}^2\right)\int_1^\infty\frac{1}{\left(\left(\text{a}-1\right)^2+\sigma^2\right)\cdot\left(\left(1+\text{a}\right)^2+\sigma^2\right)}\space\text{d}\sigma\tag7$$