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The value of the integral $\displaystyle \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx$ is?

The value of the integral $\displaystyle \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy$

I couldn't figure out the trick here. Looks like both of them are similar. Please help if anyone.

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    Please don't down vote . I really don't know how to write in symbols so I uploaded the PIC. And to save time I posted multi questions – 2017-01-23
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    Please remember, Math.SE is **not** a homework site. You should provide extra context about your attempts on the problem and try to be less rude by saying "solve it.", and instead try to ask for help, politely. – 2017-01-23
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    I understand that you may have literally no idea how to solve it. You can, however, provide your background so as to convince others that you are really in trouble as well as allow answer-er adjust the level of the answer. Simply throwing a question seems to me the best way to make readers unmotivated. – 2017-01-23
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    And here are some hints: For the first integral, applying Fubini's theorem to interchange the order of integration does the job. For the second one, it is a lot easier if you are aware of some basic knowledge on quadratic forms (or on multivariate normal distribution). You can simply diagonalize the quadratic form. Even if you have not heard of these topics, you can complete the square and resolve each integral at once, using the well-known gaussian integral formula $$ \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}, \qquad a > 0. $$ – 2017-01-23
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    The first one as suggested above, The second one, take a look at the standard normal distribution – 2017-01-23
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    I am an undergraduate preparation for entrance exam. I have not seen any such questions before. Iam in a hurry as my doubts are piling up . so please bear with me. – 2017-01-23

1 Answers 1

2

Here are solutions.

Solution of 1. Apply the Fubini's theorem to interchange the order of integration. The domain of integration is specified by the inequality $0 \leq x \leq y$, which tells you that

$$ \int_{0}^{\infty}\int_{x}^{\infty} \frac{e^{-y}}{y} \, dydx = \int_{0}^{\infty}\int_{0}^{y} \frac{e^{-y}}{y} \, dxdy = \int_{0}^{\infty} e^{-y} \, dy = 1. $$

Solution of 2. There are severay ways of solving this. Let me first show you a solution using the gaussian integral:

$$ \int_{-\infty}^{\infty} e^{-ax^2} \, dx = \sqrt{\frac{\pi}{a}}, \qquad a > 0. \tag{*} $$

Completing the square, you have

$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3(x+\frac{\sqrt{2}}{3}y)^2 - \frac{7}{3}y^2} \, dxdy. $$

Integrating w.r.t. $x$ first, we may apply the substitution $x+\frac{\sqrt{2}}{3}y = \tilde{x}$. Then

\begin{align*} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-3\tilde{x}^2 - \frac{7}{3}y^2} \, dxdy \\ &= \sqrt{\frac{\pi}{3}} \cdot \sqrt{\frac{3\pi}{7}} \\ &= \frac{\pi}{\sqrt{7}}. \end{align*}

Solution of 2, ver.2. Let me give an alternative answer which does not require the knowledge on $\text{(*)}$. Using the polar coordinates,

\begin{align*} &\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{-(3x^2+2\sqrt{2}xy + 3y^2)} \, dxdy \\ &= \int_{-\pi}^{\pi}\int_{0}^{\infty} e^{-r^2(3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta)} \, rdrd\theta \\ &= \frac{1}{2} \int_{-\pi}^{\pi} \frac{1}{3\cos^2 \theta + 2\sqrt{2}\cos\theta\sin\theta + 3\sin^2 \theta} \, d\theta \\ &= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\sec^2\theta}{3+ 2\sqrt{2}\tan\theta + 3\tan^2 \theta} \, d\theta \\ &= \int_{-\infty}^{\infty} \frac{dt}{3+ 2\sqrt{2}t + 3t^2}. \qquad (t = \tan\theta) \end{align*}

This integral can be evaluated multiple ways.