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 December 26th, 2016, 04:27 AM #1 Newbie   Joined: Dec 2016 From: USA Posts: 2 Thanks: 0 domain of a function Hi, can you help me to solve this question: find the domain of definition of the function $\displaystyle f(x)=\int_{1}^{5}{\frac{dt}{\cos(tx\sqrt{3})+\sin( x+t)}}$ Thanks. Last edited by skipjack; December 27th, 2016 at 04:09 PM.
 December 26th, 2016, 02:47 PM #2 Global Moderator   Joined: May 2007 Posts: 6,496 Thanks: 579 Since sin and cos are defined for all real arguments, it appears that x can be any real number. Thanks from ProofOfALifetime
 December 26th, 2016, 04:47 PM #3 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus Here is an idea I came up with. Since division by zero is not allowed you need to find out when the denominator of the integrand is zero. Those points will not be part of the domain. Therefore, you need to find when $\cos(tx\sqrt{3})+\sin(t+x)=0$. We know this can be true since $\cos\left(k\pi-\frac{\pi}{4}\right)+\sin\left(k\pi-\frac{\pi}{4}\right)=0$ when $k\in\Bbb{Z}$. This creates a system of equations: $\displaystyle \sqrt{3}tx=k\pi-\frac{\pi}{4}$ $\displaystyle t+x=k\pi-\frac{\pi}{4}$. Then you need to find values of $k$ which results in $t\in[1,5]$ when the system of equations is solved. Finally, you have some values of x that are not in the domain. That's not all of them, though, since it is also true that $\cos\left(k\pi\right)+\sin\left(k\pi-\frac{\pi}{2}\right)=0$. I think the domain of $f(x)$ is empty because sine eventually takes on the opposite value of what cosine is, regardless of what x is. Thanks from ProofOfALifetime Last edited by Compendium; December 26th, 2016 at 05:44 PM.
 December 27th, 2016, 01:16 PM #4 Newbie   Joined: Dec 2016 From: USA Posts: 2 Thanks: 0 thanks for help!!!
December 27th, 2016, 01:57 PM   #5
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The structure is actually pretty complex. There are lots of places where the denominator is 0.

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 December 27th, 2016, 03:56 PM #6 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus Exactly. And if you plot f(x) itself, you can see that it is defined nowhere. It also occurred to me that if $x=0$, $\cos(\sqrt{3}tx)=1$, which sine takes on the opposite once when $t\in(1,5)$. As $|x|$ increases, the period of cosine increases, meaning there will be more points where the denominator of the integrand is zero. Therefore, the integral passes through a singularity no matter what. This is assuming $x\in\Bbb{R}$. Last edited by skipjack; December 27th, 2016 at 04:11 PM.
 January 9th, 2017, 01:01 AM #7 Senior Member     Joined: Sep 2007 From: USA Posts: 349 Thanks: 67 Math Focus: Calculus I have been poking around Maxima's advanced plotting options and decided to make an animation. Here is a plot of $y=\cos(tx\sqrt{3})+\sin(t+x)$ as x changes on the interval $x\in[-3,3]$. As you can see, the integral passes through a singularity regardless of what x is. Last edited by Compendium; January 9th, 2017 at 01:06 AM.

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