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I am trying to show that this integral $$\int_0^\infty \int_0^\infty \frac{\sin \pi x}{(y+e^x|\sin \pi x|)^2}dx \, dy $$ exists and is finite and then finding its value.

Since $\sin \pi x$ takes positive and negative values, Tonelli does not apply and hence we should use Fubini. To use Fubini we need to prove that $f(x,y)=\sin \pi x/(y+e^x|\sin\pi x|)^2$ is integrable with respect to $m\times m$ Lebesgue product measure on $[0,\infty)\times [0,\infty)$. Notice that \begin{align*} \int_{[0,\infty)\times [0,\infty)}\dfrac{|\sin \pi x|}{(y+e^x |\sin \pi x|)^2}\,dm(x\times y)&=\int_{[0,\infty)\times [0,\infty)}\dfrac{1}{(\dfrac{y}{\sin^2\pi x}+\dfrac{e^x}{|\sin\pi x|})^2}\,dm(x\times y)\\ &\leq \int_{[0,\infty)\times [0,\infty)}\dfrac{1}{(y+e^x)^2}\,dm(x\times y) \end{align*} How can I prove that $f(x,y)$ is integrable? Any other ideas? Maybe we can use Tonellis somehow if we find a region such that $\sin \pi x$ is positive and the double integral of $f(x,y)$ out of this region is $0$?

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I will here focus on finding the value of the integral leaving out the justification of all the steps (which should be filled in). We first switch the order of the integration (this requires Fubini) and integrate over $y$ with the result

$$I = \int_0^\infty \left[-\frac{\sin(\pi x)}{y + e^x|\sin(\pi x)|}\right]_{y=0}^\infty dx = \int_0^\infty \frac{\sin(\pi x)}{|\sin(\pi x)|}e^{-x} dx$$

To evalute the last integral we split $[0,\infty)$ into intervals where $\frac{\sin(\pi x)}{|\sin(\pi x)|} \equiv \text{sign}(\sin(\pi x))$ is constant. This gives

$$I = \int_0^\infty \frac{\sin(\pi x)}{|\sin(\pi x)|}e^{-x} dx = \sum_{n=0}^\infty \int_n^{n+1}(-1)^{n}e^{-x}dx = \sum_{n=0}^\infty (e^{-n}-e^{-(n+1)})(-1)^{n}$$

which is just a geometric series with sum

$$I = \frac{e-1}{e+1}$$

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  • $\begingroup$ Finding the integral is easy, but the question is how to make Fubini theorem or Tonelli valid. $\endgroup$ – Ruzayqat Dec 25 '14 at 20:16
  • $\begingroup$ @Leo You write "...and then finding its value...". If you already know this you should state so in the question so that people don't waste time answering something you already know. $\endgroup$ – Winther Dec 25 '14 at 20:17
  • $\begingroup$ Of course thanks for your answer and your help which is very appreciated. I didn't mean to be rude. $\endgroup$ – Ruzayqat Dec 25 '14 at 20:20
  • $\begingroup$ I think I found the answer by applying Tonellis to $1/(y+e^x)^2$ because it is positive and measurable. So, we need to Tonellis to prove that Fubini applies. $\endgroup$ – Ruzayqat Dec 25 '14 at 20:42
  • $\begingroup$ @Leo If you have solved your own problem then you can (and should try) to post an answer to your own question. You can also accept your own answer. This makes it easier for people that comes later and has a similar question:) $\endgroup$ – Winther Dec 25 '14 at 20:44
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Notice that $f(x,y)=1/(y+e^x)^2$ is continuous for all $x,y\in [0,\infty)$, hence $m\times m$-Lebesgue measurable, $f(x,y)\geq 0$ and $([0,\infty),\mathcal{M},m)$ is $\sigma$-finite measure space. Thus by Tonelli's Theorem \begin{align*} \int_{[0,\infty)\times [0,\infty)}\dfrac{1}{(y+e^x)^2}\,dm(x\times y)&=\int_0^\infty\int_0^\infty \dfrac{1}{(y+e^x)^2}\,dy\,dx\\ &=\int_0^\infty \dfrac{-1}{y+e^x}\Big|_0^\infty \,dx=\int_0^\infty e^{-x}\,dx=1<\infty. \end{align*} Hence the function $\sin\pi x/(y+e^x|\sin\pi x|)^2$ is integrable and hence Fubini theorem applies. Then continue as Winther did.

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  • $\begingroup$ Nice additional information! +1. $\endgroup$ – Olivier Oloa Dec 26 '14 at 9:42

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