The closed-form of $\int_0^\infty \frac{e^{-x}\cos(x)}{x^2+1}\mathrm{d}x$

Question

What is the closed-form of the following integral $$I\;=\; \int_0^\infty \frac{e^{-x}\cos(x)}{x^2+1}\mathrm{d}x$$ If we replaced $\;\;\displaystyle \frac{1}{x^2+1}\;\;$ by its integral representation, $\;\;\displaystyle \int_0^\infty e^{-xt}\sin(t)\text{d}t,\;\;$ we get that $$I\;=\; \int_0^\infty e^{-x}\cos(x)\Bigg(\int_0^\infty e^{-xt}\sin(t)\text{d}t\Bigg)\mathrm{d}x$$ $$ \{\text{ reverse the order of integration } \} $$ $$I\;=\; \int_0^\infty \sin(t) \Bigg(\int_0^\infty e^{-x(t+1)}\cos(x)\text{d}x\Bigg)\mathrm{d}t$$ $$I\;=\; \int_0^\infty \frac{(t+1)\sin(t)}{(t+1)^2+1} \mathrm{d}t$$ $$\{\text{ make the change of variable $t+1=u$}\big\}$$ $$=\int_1^\infty \frac{u\sin(u-1)}{u^2+1} \mathrm{d}t $$ $$=\;\;\cos(1)\int_1^\infty \frac{u\sin(u)}{u^2+1} \mathrm{d}t\;-\;\sin(1)\int_1^\infty \frac{u\cos(u)}{u^2+1} \mathrm{d}t $$ Have anyone an idea to finish the remaining integrals ?

Answer 1

The Laplace transform of ${\displaystyle \frac{1}{1+x^{2}}}$ is well known $$\int_{0}^{+\infty}\frac{e^{-sx}}{1+x^{2}}dx=\mathrm{Ci}\left(s\right)\sin\left(s\right)+\frac{\pi\cos\left(s\right)}{2}-\mathrm{Si}\left(s\right)\cos\left(s\right)$$ where $\mathrm{Ci}\left(s\right),\,\mathrm{Si}\left(s\right)$ is the cosine and sine integral (see here), then note that $$\int_{0}^{+\infty}\frac{e^{-x}\cos\left(x\right)}{1+x^{2}}dx=\mathrm{Re}\left(\int_{0}^{+\infty}\frac{e^{-x+ix}}{1+x^{2}}dx\right)$$ $$=\color{red}{\mathrm{Re}\left(\mathrm{Ci}\left(1-i\right)\sin\left(1-i\right)+\frac{\pi\cos\left(1-i\right)}{2}-\mathrm{Si}\left(1-i\right)\cos\left(1-i\right)\right)}\approx 0.47941$$ and I don't think we can be more explicit than that.

Answer 2

Well maybe another way to attack this, we have the following integral:

$$\mathcal{I}_\text{n}:=\int_0^\infty\frac{\exp\left(-x\right)\cos\left(\text{n}x\right)}{1+x^2}\space\text{d}x\tag1$$

Now, find:

$$\mathcal{I}_\text{n}''=\frac{\text{d}^2\mathcal{I}_\text{n}}{\text{dn}^2}=\int_0^\infty\frac{\partial^2}{\partial\text{n}^2}\left(\frac{\exp\left(-x\right)\cos\left(\text{n}x\right)}{1+x^2}\right)\space\text{d}x=-\int_0^\infty\frac{x^2\exp\left(-x\right)\cos\left(\text{n}x\right)}{1+x^2}\space\text{d}x\tag2$$

Now, we can write:

$$\mathcal{I}_\text{n}+\mathcal{I}_\text{n}''=\int_0^\infty\frac{\left(1+x^2\right)\exp\left(-x\right)\cos\left(\text{n}x\right)}{1+x^2}\space\text{d}x=\int_0^\infty\exp\left(-x\right)\cos\left(\text{n}x\right)\space\text{d}x\tag3$$

Using the well-known result of the Laplace transform:

$$\mathcal{I}_\text{n}+\mathcal{I}_\text{n}''=\frac{1}{1+\text{n}^2}\tag4$$

But the necessary boundary conditions become nasty very quickly.