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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2You can evaluate the integral in terms of the exponential integral. – Mhenni Benghorbal Apr 03 '13 at 12:33
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I know but I don't consider that elementary. I also don't know enough about the exponential integral to like using it. – Ishan Banerjee Apr 03 '13 at 12:35
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6@IshanBanerjee: Not every integral can be evaluated in terms of elementary functions. – Mhenni Benghorbal Apr 03 '13 at 12:38
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1The indefinite integral isn't elementary , but is it possible to do one with these limits without involving the exponential integral? – Ishan Banerjee Apr 03 '13 at 12:41
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1Nope. 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. ain't a mathematician Apr 03 '13 at 12:42
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@J.M: The Laplace transform of this function is in terms of the exponential function. Here is the Laplace transform $$ 1/2,i \left( {{\rm e}^{is}}{\it Ei} \left( 1,is \right) -{{\rm e}^{-i s}}{\it Ei} \left( 1,-is \right) \right) . $$ – Mhenni Benghorbal Apr 03 '13 at 12:45
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Not quite, @Mhenni. Exponential integral maybe... – J. M. ain't a mathematician Apr 03 '13 at 12:46
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@J.M. I did and found stuff involving Ci and Si. – Ishan Banerjee Apr 03 '13 at 12:48
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Ishan, quite right. Those functions are related to $\mathrm{Ei}$ in much the same way $\cos$ and $\sin$ are related to $\exp$; hence @Mhenni's first comment. – J. M. ain't a mathematician Apr 03 '13 at 12:49
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@IshanBanerjee: These are the sine and cosine integrals and they are related to exponential integrals. – Mhenni Benghorbal Apr 03 '13 at 12:49
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So, what do I do now? Do I delete the question, seeing as they can't be solved using only elementary functions? – Ishan Banerjee Apr 03 '13 at 12:53
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1@IshanBanerjee: Why should you delete it? Just leave it. – Mhenni Benghorbal Apr 03 '13 at 12:58
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@IshanBanerjee: It is a good idea to have a close look at some special functions to be able to handle some of these non elementary integrals. – Mhenni Benghorbal Apr 03 '13 at 12:59
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One can use Chebyshev polynomials for $\frac{1}{1+x^2}$ and then the integrals are computable. But then one has to cope with a series that can also have convergence problems. – Jon Apr 10 '13 at 08:04
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Can you elaborate on this ? – Jean-Claude Arbaut Apr 11 '13 at 14:12
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Taylor expanding the denominator and then integrating term-by-term gives a divergent series: $$\int_0^\infty \frac{e^{-x}}{1+x^2} = \sum_{n=0}^\infty (-1)^n(2n)!.$$ I wonder if this can be converted to a convergent series. – Douglas B. Staple Apr 11 '13 at 14:40
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@arbautjc Referring to the link ,the integral on the big arc doesn't tend to 0 for this case. – Ishan Banerjee Apr 11 '13 at 15:33
2 Answers
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As pointed out in the comments and in Xiaolang's answer, this integral is very unlikely to have a closed-form solution in terms of elementary functions, but it is possible to express it in terms of exponential integrals or, equivalently, sine and cosine integrals. These hardly qualify as "elementary methods", but we can at least eliminate the sine and cosine integrals using their power series expansions.
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.
Douglas B. Staple
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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 02:58
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1Though we didn't get an elementary answer, your post provided some additional insight. Congratulations to the +100. – Mårten W Apr 12 '13 at 11:08
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@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 02:43
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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
but you can get a approximation (WolframAlpha can help you)
$$\int_0^{\pi/2}e^{\tan(\theta)}d\theta= \operatorname{Ci}(1)\sin(1)- \operatorname{Si}(1)\cos(1)+\pi \cos(1)/2 \approx 0.62144962423581335763926$$
