# $\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx$

which ways to compute this integral:

$$\int_{-\infty}^{\infty} e^{tx} \frac{1}{\pi(1+x^2)} \mathrm dx$$

for different cases of t. t can be any value in $\mathbb{R}$. I was stuck at how to get the antiderivative of the integrand. Thank you!

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The integrand probably doesn't have a closed-form antiderivative. But the usual contour integral method works. Are you familiar with complex analysis? – Qiaochu Yuan Dec 12 '10 at 2:45
Sorry, I am not. But really appreciate if you can post you way. I will try to understand – Jessie Dec 12 '10 at 2:47
Yes, the antiderivative isn't elementary (the exponential integral isn't elementary, 'no?). Are you a beginning calculus student or a professional? – J. M. Dec 12 '10 at 2:50
I don't think it is a good idea to remove so much of your question at this point. Zaricuse's and Shai Covo's answers address the integral you removed. For the remaining integral, as mentioned in my answer you can use $\arctan$ for the only case where it converges ($t=0$). – Jonas Meyer Dec 12 '10 at 19:44
$\large t > 0$ ? – Felix Marin Jan 28 '14 at 10:08

When $t \neq 0$, the first integral diverges and the second one converges only if $t < 0$. In the latter case, if you call your integral $f(t)$, then differentiating under the integral sign gives that $$f''(t) + f(t) = \int_0^{\infty} {e^{tx} + x^2e^{tx} \over \pi(1 + x^2)}$$ $$= {1\over \pi} \int_0^{\infty}e^{tx}$$ The latter integral integrates to ${\displaystyle-{1 \over t}}$, so that you have ${\displaystyle f''(t) + f(t) = -{1 \over \pi t}}$. This can be solved via variation of parameters, and (according to wolframalpha at least) it does not have an elementary expression.

On the other hand the nonelementary functions in the solution just involve indefinite integrals of ${\displaystyle {\sin(t) \over t}}$ and ${\displaystyle{\cos(t) \over t}}$, so if you want to do something this advanced you might be able to get something satisfactory using initial conditions for this differential equation.

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First, check whether they converge. You will find that the first one doesn't unless $t=0$, and the second one doesn't unless $t\leq0$. For the first integral, when $t=0$, you can use the arctangent function. For the second, when $t\lt0$, you can try to use complex contour integration methods as seen in many texts on complex analysis.

Edit: I was hasty in posting this, and hadn't actually thought through how the contour integration would go. Based on the other answers, it looks like it wouldn't be so straightforward.

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Often these easier contour integrals can be done using differentiation under the integral sign. I think one has to introduce an additional parameter to do this, though. – Qiaochu Yuan Dec 12 '10 at 2:55
@Qiaochu: That's interesting, I didn't know that. Unless you're already planning on posting an answer, could you perhaps give a reference for an example of that method? – Jonas Meyer Dec 12 '10 at 2:59
@Jonas: see en.wikipedia.org/wiki/… . – Qiaochu Yuan Dec 12 '10 at 3:01
@Jonas:Thanks! what theorems can I use to check their convergence? In the link, I saw the example where the nominator is e^{itx}. That is complex number, so isn't it different from my integrands? – Jessie Dec 12 '10 at 3:22
@Qiaochu Yuan: if possible, could you elaborate your method? – Jessie Dec 12 '10 at 3:24

Correspondingly to Zaricuse's second paragraph, according to this (see the formula before the last one), for any $s > 0$, $$\int_0^\infty {e^{ - sx} \frac{1}{{\pi (1 + x^2 )}}\,{\rm d}x} = \frac{1}{\pi } \Big \lbrace \cos (s)\Big[\frac{\pi }{2} - {\rm Si}(s)\Big] - \sin (s){\rm Ci}(s)\Big \rbrace,$$ where ${\rm Si}(s) = \int_0^s {\frac{{\sin u}}{u}} \,{\rm d}u$ is the Sine Integral, and ${\rm Ci}(s) = \int_s^\infty {\frac{{\cos u}}{u}\,{\rm d}u}$ the Cosine Integral.

EDIT: Probabilistic interpretation of the integrals.

The function $f(x) = \frac{1}{{\pi (1 + x^2 )}}$, $x \in \mathbb{R}$, is the density function of a standard Cauchy random variable $X$. The divergent integral $${\rm E}[e^{tX}] = \int_{ - \infty }^\infty {e^{tx} f(x)\,{\rm d}x} = \infty, \;\; t \neq 0,$$ corresponds exactly to the (elementary) fact that the moment-generating function of the Cauchy distribution does not exist. On the other hand, the characteristic function is well known to be $${\rm E}[e^{{\rm i}{\rm t}X}] = \int_{ - \infty }^\infty {e^{{\rm i}tx} f(x)\,{\rm d}x} = e^{ - |t|}, \;\; t \in \mathbb{R}.$$ Finally, the function $\tilde f(x) = \frac{2}{{\pi (1 + x^2 )}}$, $x > 0$, is the density function of $|X|$. Then, the Laplace transform of $|X|$ is $${\rm E}[e^{-t|X|}] = \int_0^\infty {e^{ - tx} \tilde f(x)\,{\rm d}x},\;\; t > 0,$$ and, according to the first paragraph, can be expressed in terms of the Sine Integral and Cosine Integral functions.

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Yes, this is what I got as well; I'll make the notational note that sometimes the function $\mathrm{si}(x)=\frac{\pi}{2}-\mathrm{Si}(x)$ is used to express this instead. – J. M. Dec 12 '10 at 9:23
(Alternatively, one can express this in terms of exponential integrals of imaginary argument, similar to being able to express sines and cosines in terms of complex exponentials.) – J. M. Dec 12 '10 at 9:57