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How can I solve this integral: $$\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx.$$ Can I solve this problem using the Laplace transform? How can I do this?

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On that note, somebody was asking about the Fourier-Hermite expansion of $\dfrac1{1+x^2}$ a few days ago. Are you the same guy? –  J. M. Dec 23 '11 at 23:12
    
Based on the Gravatar, you apparently are one and the same... –  J. M. Dec 24 '11 at 0:58
    
Thank you all for your help to solve this problem! –  using Dec 24 '11 at 4:33
    
@using, you're welcome! If one of the solutions is satisfactory (cough cough), you can accept it as the official solution by clicking the check next to the vote count. :-) –  Bruno Joyal Dec 24 '11 at 19:17
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5 Answers

up vote 16 down vote accepted

I will evaluate the integral

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

using one of my favorite techniques (which I have heard was also Richard Feynman's favorite). This will solve your problem since, as Dilip points out, your integral can be written as

$$\int_{-\infty}^\infty\frac{e^{-x^2}(2x^2+2-3)}{1+x^2}dx=2\sqrt{\pi}-3I.$$

To find the value of $I$, we let, for $t\geq 0$,

$$I(t)=\int_{-\infty}^\infty\frac{e^{-tx^2}}{1+x^2}dx.$$

Then $I(0)=\arctan{\infty}-\arctan{(-\infty)} = \pi$, and $$I'(t)=\int_{-\infty}^\infty\frac{-x^2e^{-tx^2}}{1+x^2}dx.$$

Hence $I(t)$ satisfies the differential equation

$$I(t)-I'(t)=\int_{-\infty}^\infty e^{-tx^2}dx = \sqrt{\frac{\pi}{t}}.$$

Multiplying throughout by $e^{-t}$ we have

$$-\frac{d}{dt}(e^{-t}I(t)) = e^{-t}\sqrt{\frac{\pi}{t}}.$$

Integrating from $t=0$ to $t$, we find

$$-e^{-t}I(t)+I(0) = \sqrt{\pi} \int_0^t \frac{e^{-t}}{\sqrt t}dt = 2\sqrt{\pi} \int_0^\sqrt{t}e^{-u^2} du = \pi \text{ erf}\sqrt{t}$$

Since $I(0)=\pi$, we have

$$I(t)=e^t(\pi - \pi \text{ erf}(\sqrt{t})) = e^{t}\pi \text{ erfc}(\sqrt{t}).$$

Thus $I=I(1) = e\pi\: \text{erfc}(1)$, and your integral is $2\sqrt\pi - 3e\pi\text{erfc}(1)$. (Look!)

Note: my other solution below is much quicker, but not as much fun.

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Ah, this is what I wanted to see... +1! I suppose integration by parts will take care of the higher moments $\int_{-\infty}^\infty\frac{e^{-x^2}}{1+x^2}x^{2k}\mathrm dx$ –  J. M. Dec 24 '11 at 0:06
    
Thanks @J.M.! :-) –  Bruno Joyal Dec 24 '11 at 0:08
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Here is yet another way to evaluate the integral

$$I=\int_{-\infty}^\infty \frac{e^{-x^2}}{1+x^2}dx.$$ Write

$$\frac{1}{1+x^2}= \int_0^\infty e^{-s(1+x^2)}ds,$$

so that

$$I=\int_{-\infty}^\infty e^{-x^2} \int_{0}^\infty e^{-s(1+x^2)}ds\: dx.$$

We switch the order of integration to get

$$\int_{0}^\infty e^{-s} \int_{-\infty}^\infty e^{-x^2(1+s)}dx\: ds = \int_{0}^\infty e^{-s} \sqrt{\frac{\pi}{1+s}} ds = e\sqrt{\pi}\int_{0}^\infty \frac{e^{-(s+1)}}{\sqrt{1+s}} ds = e\pi \text{ erfc}(1).$$

(To see that this last integral is indeed $\sqrt\pi \text{ erfc}(1)$, put $u=\sqrt{1+s}$, $du= \frac{ds}{2\sqrt{1+s}}$.)

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I liked the other way (which I upvoted) better because it does not require looking up (or knowing) that that last integral has value $\sqrt{\pi}\text{erfc}(1)$. Ditto for @J.M.'s one-liner comment which has since been deleted. –  Dilip Sarwate Dec 24 '11 at 1:21
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Thanks Dilip! Doesn't it follow immediately from the definition of $\text{erfc}$ after the substitution $u=\sqrt{1+s}$? –  Bruno Joyal Dec 24 '11 at 1:23
    
I didn't notice that it would work out that way, but then I don't tend to use erf or erfc at all, preferring $\Phi(\cdot)$, the cumulative normal distribution instead (cf. my own answer to the problem). In fact, I request you to edit your answer to point out how that last step follows by simple substitution. Thanks –  Dilip Sarwate Dec 24 '11 at 1:26
    
You don't have to look things up if you know how the complementary error function $\mathrm{erfc}(x)=\int_x^\infty \exp(-u^2)\mathrm du$ is defined. Let $s=u^2-1$, $\mathrm ds=2u\mathrm du$. –  J. M. Dec 24 '11 at 1:26
    
@Dilip: I see why you're used to the normal CDF, so I understand why you like that route. Me, I like being able to exploit the oddness of the error function, and a few other things besides. –  J. M. Dec 24 '11 at 1:28
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To find $I=\int_{-\infty}^{\infty}\frac{e^{-x^2}(2x^2-1)}{1+x^2}dx$, let us start by defining for $n \in \mathbb{N}$ $$ I_n = \int_{-\infty}^{\infty} x^{2n} e^{-x^2} $$ and note that $I_0=\sqrt{\pi}$, also called the Gaussian integral, can be solved either by a beautiful polar coordinate trick or by this technique, and that $I_1=\frac{\sqrt{\pi}}{2}$ was shown here. We can find the rest using integration by parts. With $$ \begin{array}{lcl} v=-\frac{1}{2}e^{-x^2} && dv=xe^{-x^2}dx \\ u=x^{2n-1} && du=(2n-1)x^{2n-2}dx, \end{array} $$ for $n>1$ we have $$ \begin{array}{lcl} I_n &=& \int udv = uv - \int vdu \\ &=& -\frac{1}{2} \left[ x^{2n-1} e^{-x^2} \right]_{-\infty}^{\infty} + \frac{2n-1}{2} \int_{-\infty}^{\infty} x^{2n-2} e^{-x^2} \\ &=& \frac{2n-1}{2} I_{n-1} \end{array} $$ where the first right-hand term vanishes because the exponential term dominates all polynomial terms. This shows that, inductively, $$ I_n = \frac{(2n)!}{2^{2n}n!} I_0 $$ But using $\frac{1}{1-t}=\sum_{n=0}^{\infty}t^n$, we can expand the integrand of $I$: $$ \frac{2x^2-1}{1+x^2} = (2x^2-1) \sum_{n=0}^{\infty} (-1)^n x^{2n} = 2-3\sum_{n=0}^{\infty} (-1)^n x^{2n} $$ and hence we get (noting that the integrals all converge because of the dominating exponential term) $$ \begin{array}{lcl} I &=& 2I_0-3(I_0-I_1+\dots) = 2I_0-3 \sum_{n=0}^{\infty} (-1)^n I_n \\ &=& \left( 2-3 \sum_{n=0}^{\infty} (-1)^n \frac{(2n)!}{2^{2n}n!} \right) I_0 \end{array} $$ where the series $$ \sum_{n=0}^{\infty} (-1)^n \frac{(2n)!}{2^{2n}n!} = 1 - \frac{1}{2} + \frac{1 \cdot 3}{2 \cdot 2} - \frac{1 \cdot 3 \cdot 5}{2 \cdot 2 \cdot 2} + \cdots = e \sqrt\pi\; \text{erfc}(1) $$ can also be expressed using $\Gamma(\frac{1}{2}-n)=(-1)^n\frac{(2n)!}{4^{n}n!}\sqrt\pi$ and a series expansion for the complementary error function as $$ I = 2\sqrt\pi - 3 \; \sum_{n=0}^{\infty} \; \Gamma(\tfrac{1}{2}-n) = 2\sqrt\pi - 3e\pi\;\text{erfc}(1) \;. $$

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Why are you allowed to use the geometric series? I don't understand why we can guarantee $|t|<1$. –  ae0709 Dec 23 '11 at 22:08
    
The geometric series doesn't converge outside the radius of convergence you give. But when we multiply by the exponential in the next step, each term converges since $x^ne^{-x} \to 0$ (and so also $x^ne^{-x^2} \to 0$) as $x\to\infty$ for all $n\in\mathbb{N}$. –  bgins Dec 23 '11 at 22:32
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Your last series needs some love. –  Jonas Teuwen Dec 23 '11 at 22:33
    
Yes, you caught me between edits, with a nasty mistake! –  bgins Dec 23 '11 at 23:19
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Writing $2x^2-1$ as $2x^2 + 2 - 3$, the integral simplifies to $$\begin{align*} \int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx &= 2\int_{-\infty}^{\infty}\exp(-x^2)\mathrm dx -3 \int_{-\infty}^{\infty}\exp(-x^2)\frac{1}{1+x^2}\mathrm dx\\ &= 2\sqrt{\pi} - \frac{3}{2\pi}\int_{-\infty}^{\infty}\sqrt{\pi}\exp(-\omega^2/4) \pi\exp(-|\omega|)\mathrm d\omega\\ &= 2\sqrt{\pi} - 3\sqrt{\pi}\int_{0}^{\infty}\exp(-\omega^2/4) \exp(-\omega)\mathrm d\omega\\ &= 2\sqrt{\pi} - 6\pi e \int_{0}^{\infty}\frac{1}{\sqrt{2}\sqrt{2\pi}} \exp\left(-\frac{(\omega + 2)^2}{2\cdot(\sqrt{2})^2}\right) \mathrm d\omega\\ \end{align*}$$ where in the second step, we have used a well-known result (see e.g. the answer by bgins and the links therein) on the first integral, and applied the inner-product form of Parseval's theorem to convert the second integral to the integral of the product of the Fourier transforms $\sqrt{\pi}\exp(-\omega^2/4)$ and $\pi\exp(-|\omega|)$ of $\exp(-x^2)$ and $(1+x^2)^{-1}$ respectively, while the last step follows upon completing the square in the exponent and writing the integrand as the probability density function of a normal random variable with mean $-2$ and variance $2$. Hence we have that $$\begin{align*} \int_{-\infty}^{\infty}\exp(-x^2)\frac{2x^2-1}{1+x^2}\mathrm dx &= 2\sqrt{\pi} -6\pi e \Phi(-2/\sqrt{2})\\ &= 2\sqrt{\pi} -6\pi e Q(\sqrt{2}) \end{align*}$$ where $\Phi(\cdot)$ is the cumulative standard normal distribution function and $Q(x) = 1-\Phi(x)$ is its complement. Since $\text{erfc}(x) = 2Q(x\sqrt{2})$, the value can also be expressed as $2\sqrt{\pi} - 3e\pi \text{erfc}(1)$ as in the answers by Bruno and bgins.

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Would the down-voter care to add a comment? –  Dilip Sarwate Mar 7 '12 at 13:15
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Hint: If you want to use Laplace transform, you'll want a change of variables $x^2 = t$.

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Why the downvotes? Using that change of variables (and symmetry) the integral becomes $\int_0^\infty \frac{2t-1}{(t+1)\sqrt{t}} e^{-t}\ dt= 2 \int_0^\infty \frac{1}{\sqrt{t}} e^{-t}\ dt - 3 \int_0^\infty e^{-t} \frac{1}{\sqrt{t} (1+t)}\ dt$. Any decent table of Laplace transforms will handle this. –  Robert Israel Dec 25 '11 at 4:27
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