# Evaluating $\int_0^{\infty}\frac{e^{-x}}{1+x^2}dx$

I'm trying to evaluate $$\int_0^{\infty}\dfrac{e^{-x}}{1+x^2}dx$$ By making the substitution $x=\tan\theta$, $$\int_0^{\infty}\dfrac{e^{-x}}{1+x^2}dx=\int_0^{\frac \pi 2}\exp(-\tan\theta)d\theta$$ So it converges to something less than $\frac \pi 2$. Is there any way to find the exact value, using only elementary methods?

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You can evaluate the integral in terms of the exponential integral. – Mhenni Benghorbal Apr 3 '13 at 12:33
@IshanBanerjee: Not every integral can be evaluated in terms of elementary functions. – Mhenni Benghorbal Apr 3 '13 at 12:38
The indefinite integral isn't elementary , but is it possible to do one with these limits without involving the exponential integral? – Ishan Banerjee Apr 3 '13 at 12:41
Nope. Look up the Laplace transform of $\frac1{1+x^2}$ in any convenient table (or evaluate with any software at hand), and report back. – J. M. Apr 3 '13 at 12:42
@IshanBanerjee: Why should you delete it? Just leave it. – Mhenni Benghorbal Apr 3 '13 at 12:58

there may not be a elementray functions but it isn't a easy thing to prove you can consult The Liouville's therom in complex analysis

$$\int_0^{\pi/2}e^{\tan(\theta)}d\theta= \operatorname{Ci}(1)\sin(1)- \operatorname{Si}(1)\cos(1)+\pi \cos(1)/2 \approx 0.62144962423581335763926$$
We start with the representation in terms of the cosine and sine integrals $\operatorname{Ci}$ and $\operatorname{Si}$: \begin{align} \int_0^\infty \frac{e^{-x}}{1+x^2} &= \operatorname{Ci}(1)\sin(1) - \operatorname{Si}(1)\cos(1)+\frac{\pi \cos(1)}{2},\\\ \\ \operatorname{Ci}(x) &= \gamma + \ln(x) +\int_0^x \frac{\cos(t)-1}{t}dt, \\\ \\ \operatorname{Si}(x) &= \int_0^x \frac{\sin(t)}{t}dt, \end{align} where $\gamma=0.577\ldots$ is the Euler-Mascheroni constant. Power series expansions for $\operatorname{Ci}(x)$ and $\operatorname{Si}(x)$ follow directly from their definitions in terms of sines and cosines: \begin{align} \operatorname{Ci}(x) &= \gamma + \ln(x) + \sum_{j=0}^\infty \frac{(-1)^{j+1}x^{2j+2}}{(2j+2)!(2j+2)},\\\ \\ \operatorname{Si}(x) &= \sum_{j=0}^\infty \frac{(-1)^jx^{2j+1}}{(2j+1)!(2j+1)}. \end{align} Substituting $x=1$ into these expressions, applying the power series expansions for $\sin(1)$ and $\cos(1)$ where they multiply $\operatorname{Ci}(1)$ and $\operatorname{Si}(1)$, and doing a somewhat non-negligible amount of arithmetic, we find: $$\int_0^\infty \frac{e^{-x}}{1+x^2} = \gamma \sin(1) + \frac{\pi\cos(1)}{2} + \sum_{k=0}^\infty \sum_{j=1}^\infty \frac{f(j-k)}{j(j!)(k!)}. \tag{\star}$$ Here the function $f(j-k)$ is the sequence $0,-1,0,1,\ldots$: \begin{align} f(j-k) &= 0\ \ \quad \textrm{if}\quad j-k\equiv 0\ (\operatorname{mod} 4)\\ &= -1\ \ \ \,\textrm{if}\quad j-k\equiv 1\ (\operatorname{mod} 4)\\ &= 0\ \ \quad \textrm{if}\quad j-k\equiv 2\ (\operatorname{mod} 4)\\ &= 1\ \ \quad \textrm{if}\quad j-k\equiv 3\ (\operatorname{mod} 4).\\ \end{align} Clearly Eq. $(\star)$ converges rapidly, considering the factorial functions in the denominator of the summand. Indeed, including just 25 terms (5 values for each index), we obtain 0.621..., consistent with the result 0.621449... quoted previously.
PS. In your question you quote $\pi/2$ as an upper bound. Actually, note that we have $1/(x^2+1)\leq 1\ \forall\ x\in\mathbb{R}$, with the equality only at $x=0$. This gives: $$\int_0^\infty \frac{e^{-x}}{{x^2+1}}dx < 1=\int_0^\infty e^{-x}dx.$$ – Douglas B. Staple Apr 12 '13 at 2:58
@MårtenW Thanks. I found this problem challenging and had fun looking at it. It's a shame that the double sum in $(\star)$ couldn't be identified with any known constants. I'd be interested to hear if someone manages to do better. – Douglas B. Staple Apr 14 '13 at 2:43