Convergence of $\int_{0}^{+\infty }\frac{e^{-x}} {1+x^2}dx$

Question

$$\int_{0}^{+\infty }\frac{e^{-x}} {1+x^{2}}dx$$

One more task that i got on my exam, and i failed. I tried with partial integration. I need to find out does it converge?

Answer 1

The integral converges by comparison with $\int_0^\infty e^{-x}\,\mathrm{d}x$.

Furthermore, $$ \begin{align} \int_0^\infty\frac{e^{-x}}{1+x^2}\mathrm{d}x &=\int_0^\infty\frac1{2i}\left(\frac1{x-i}-\frac1{x+i}\right) e^{-x}\,\mathrm{d}x\tag{1}\\ &=\frac1{2i}\left(\int_0^\infty\frac{e^{-x}}{x-i}\,\mathrm{d}x-\int_0^\infty\frac{e^{-x}}{x+i}\,\mathrm{d}x\right)\tag{2}\\ &=\frac1{2i}\left(e^{-i}\int_{-i}^{\infty-i}\frac{e^{-x}}{x}\,\mathrm{d}x-e^i\int_{i}^{\infty+i}\frac{e^{-x}}{x}\,\mathrm{d}x\right)\tag{3}\\ &=\frac1{2i}\left(e^{-i}\int_1^\infty\frac{e^{ix}}{x}\,\mathrm{d}x-e^i\int_1^\infty\frac{e^{-ix}}{x}\,\mathrm{d}x\right)\tag{4}\\ &=\mathrm{Im}\left(e^{-i}\int_1^\infty\frac{e^{ix}}{x}\,\mathrm{d}x\right)\tag{5}\\ &=\cos(1)\int_1^\infty\frac{\sin(x)}{x}\,\mathrm{d}x-\sin(1)\int_1^\infty\frac{\cos(x)}{x}\,\mathrm{d}x\tag{6}\\[3pt] &=\frac\pi2\cos(1)-\cos(1)\,\mathrm{Si}(1)+\sin(1)\,\mathrm{Ci}(1)\tag{7} \end{align} $$ Explanation:
$(1)$: partial fractions
$(2)$: distribute the integral over the sum
$(3)$: substitute $x\mapsto x+i$ in the first and $x\mapsto x-i$ in the second integral
$(4)$: no singularities in $[-i,R-i]\cup-i+Re^{[0,-\pi i/2]}\cup[-i-iR,-i]$
$\hphantom{(4)\text{:}}$ no singularities in $[i,R+i]\cup i+Re^{[0,\pi i/2]}\cup[i+iR,i]$
$\hphantom{(4)\text{:}}$ substitute $x\mapsto-ix$ in the first and $x\mapsto ix$ in the second integral
$(5)$: rewrite $\mathrm{Im}(z)$
$(6)$: extract the imaginary part
$(7)$: translate to special functions

Answer 2

I'll purpose a more theoretical answer.

$$\int_{0}^{+\infty }\frac{e^{-x}} {1+x^{2}} = \frac{1}{2} \int_{\mathbb{R}}\frac{e^{-|x|}} {1+x^{2}} = $$ $$ \frac{1}{2} \langle e^{-|x|}, \mathcal{F}(\sqrt{\frac{\pi}{2}}e^{-|x|})\rangle_{L^2(\mathbb{R})}\ \stackrel{C-S}{\leq} \frac{1}{2}\sqrt{\frac{\pi}{2}} \ ||e^{-|x|}||_{L^2(\mathbb{R})} ||\mathcal{F}(e^{-|x|})||_{L^2(\mathbb{R})}= \frac{1}{2}\sqrt{\frac{\pi}{2}} \sim 0.6266 $$

So we proved that: $$\int_{0}^{+\infty }\frac{e^{-x}} {1+x^{2}} \leq \frac{1}{2}\sqrt{\frac{\pi}{2}} $$

On the other hand let's observe that $1+ x^2 \leq e^x \ \forall x \geq 0$, so we get the following:

$$\int_{0}^{+\infty }\frac{e^{-x}} {1+x^{2}} \geq \int_{0}^{+\infty} e^{-x}e^{-x} = \frac{1}{2} $$

Thus we proved that: $$\frac{1}{2} \leq \int_{0}^{+\infty }\frac{e^{-x}} {1+x^{2}} \leq \frac{1}{2}\sqrt{\frac{\pi}{2}} $$

Answer 3

$$\int_0^\infty \frac{dx}{e^x (1+x^2)}=\int_{1}^\infty \frac{dt}{t^2 (1+\ln^2 t)}<\int_1^\infty \frac{dt}{t^2}<\infty.$$

Answer 4

The integrand function is non-negative on $\mathbb{R}^+$ and the Cauchy-Schwarz inequality provides an extremely tight bound for its value: $$ 0\leq I=\int_{0}^{+\infty}\frac{e^{-x}}{1+x^2}\,dx \leq \sqrt{\int_{0}^{+\infty}e^{-2x}\,dx\cdot\int_{0}^{+\infty}\frac{dx}{(1+x^2)^2}}\leq\sqrt{\frac{\pi}{8}}.\tag{1}$$ By using the Laplace transform and integration by parts we may also see that: $$ I = \int_{0}^{+\infty}\frac{\sin(s)}{s+1}\,ds=\int_{0}^{+\infty}\frac{(1-\cos(s))}{(s+1)^2}\,ds \tag{2}$$ hence: $$ I = \frac{\pi}{2}-\int_{0}^{+\infty}\frac{(1-\cos(s))(2s+1)}{s^2(s+1)^2}\,ds \tag{3}$$ and by using the Cauchy-Schwarz inequality again we have: $$ I \geq \frac{\pi}{2}-\sqrt{\int_{0}^{+\infty}\frac{(1-\cos(s))^2}{s^4}\,ds\cdot\int_{0}^{+\infty}\frac{(2s+1)^2}{(s+1)^4}\,ds} = \frac{\pi}{2}-\sqrt{\frac{7\pi}{18}}\tag{4}$$ so by putting together $(1)$ and $(4)$ we have:

$$ \color{red}{0.465}\leq\frac{\pi}{2}-\sqrt{\frac{7\pi}{18}}\leq \color{red}{I}\leq \sqrt{\frac{\pi}{8}}\leq \color{red}{0.627}\tag{5}$$

On the other hand, $(2)$ also provides the exact value of $I$ in terms of the sine and cosine integrals:

$$ \color{red}{I} = \frac{\pi}{2}\,\cos(1)+\sin(1)\,\text{Ci}(1)-\cos(1)\,\text{Si}(1)=\color{red}{0.6214496242}\ldots.\tag{6}$$

An efficient series representation follows from considering that both $\text{Si}(x)$ and $\text{Ci}(x)-(\gamma+\log(x))$ are analytic functions in a neighbourhood of zero. By computing their Taylor coefficients, we have the following improved version of $(5)$:

$$ \color{red}{I}\leq \left(\frac{\pi}{2}-\frac{17}{18}\right)\cos(1)+\left(\gamma-\frac{23}{96}\right)\sin(1)\leq \color{red}{0.6226},\\\color{red}{I}\geq \left(\frac{\pi}{2}-\frac{5109}{5400}\right)\cos(1)+\left(\gamma-\frac{1295}{5400}\right)\sin(1)\geq \color{red}{0.6214}.\tag{7}$$

Answer 5

my approach to find the value of this integral :

$$I=\int_{0}^{\infty }\frac{e^{-x}}{1+x^2}\ dx\\ \\ \\ \because \ \frac{1}{1+x^2}=\int_{0}^{\infty }sin(t)\ e^{-tx}dt\ \ \ \ \ \ \ , so \ we\ have\ \\ \\ \\ \therefore \ I=\int_{0}^{\infty }\frac{e^{-x}}{1+x^2}\ dx=\int_{0}^{\infty }\left ( \int_{0}^{\infty } e^{-tx}\ sin(t)\right )e^{-x}dx=\int_{0}^{\infty }sin(t)\left ( \int_{0}^{\infty }e^{-(1+t)x} \right )dt\\ \\ \\ =\int_{0}^{\infty }\frac{sin(t)}{1+t}dt=\int_{0}^{\infty }\frac{sin(1-1+t)}{1+t}dt=cos(1)\int_{0}^{\infty }\frac{sin(t+1)}{t+1}dt-sin(1)\int_{0}^{\infty }\frac{cos(1+t)}{1+t}dt\\ \\ \\$$

so we have $$\therefore I=cos(1)\left [ \int_{0}^{\infty }\frac{sin(t)}{t} dt-\int_{0}^{1}\frac{sin(t)}{t}\right ]-sin(1)\int_{0}^{\infty }\frac{cos(t+1)}{t+1}dt\\ \\$$

but we know that : $$\int_{0}^{1}\frac{sin(t)}{t}dt=Si(1)\ \ \ \ \ \ , \ \ \ \ \int_{0}^{\infty }\frac{cos(1+t)}{1+t}dt=-Ci(1)\ \ \ \ \ \ \ , \\ \\ \\ \ \ \ \ \ \ \ \ \ and \ \ \ \int_{1}^{\infty }\frac{cos(t)}{t}dt=Ci(1)\ \ \ \ \ \ \ , \int_{0}^{\infty }\frac{sin(t)}{t}dt=\frac{\pi }{2}\\ \\ \\ \\ \therefore \ I=sin(1)\ Ci(1)+cos(1)\ \left [ \frac{\pi }{2}-Si(1) \right ]$$