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I need to determine whether the integral

$$ \int_{-\infty}^\infty cos \,(\pi t) \;dt$$

is convergent or divergent. I rewrote this improper integral as

$$ \lim \limits_{x \to{-\infty}}\int_{x}^0 cos \,(\pi t) \;dt + \lim \limits_{x \to{\infty}}\int_0^x cos \,(\pi t) \;dt$$

I integrated $ \lim \limits_{x \to{-\infty}}\int_{x}^0 cos \,(\pi t) \;dt$ to get $ \frac{sin \,(\pi t)}{2}$ which gives me $ \frac{sin \,0}{0} - \frac{sin \,t}{t}$

Obviously I can't divide by 0 so does this mean that the function is divergent, or is there some other step I can take that I'm missing?

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    $\begingroup$ $$\lim_{x\to -\infty} \int_x^0 \cos(\pi t)\,\text{d}t = \lim_{x\to-\infty} \left(\frac{\sin(0)}{\pi} - \frac{\sin(\pi x)}{\pi}\right)$$ Note that $t$ doesn't end up in the denominator, nor is there a divide by zero error here. $\endgroup$ – Nicholas Stull Jun 11 '14 at 1:35
  • $\begingroup$ Often people will regularize integrals such as these, probably giving this one the value $0$. $\endgroup$ – Antonio Vargas Jun 11 '14 at 2:57
  • $\begingroup$ In the context of distributions $\int_{-\infty}^\infty \cos(k x)\,dx=2\pi \delta(k)$. So, in distribution, $$\int_{-\infty}^\infty \cos(\pi x)\,dx=0$$ $\endgroup$ – Mark Viola Jan 17 at 20:13
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We have $$\int_0^x \cos(\pi t)\,dt=\left[\frac{\sin(\pi t)}{\pi}\right]_0^x=\frac{\sin(\pi x)}{\pi}\ ;$$ this continues to oscillate between $1/\pi$ and $-1/\pi$ and therefore has no limit as $x\to\infty$. Hence $$\int_0^\infty \cos(\pi t)\,dt$$ diverges, and so does $$\int_{-\infty}^\infty \cos(\pi t)\,dt\ .$$

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  • $\begingroup$ In the context of distributions $\int_{-\infty}^\infty \cos(k x)\,dx=2\pi \delta(k)$. So, in distribution, $$\int_{-\infty}^\infty \cos(\pi x)\,dx=0$$ $\endgroup$ – Mark Viola Jan 17 at 20:14
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You've definitely done something wrong with that integration and substitution, if you're getting (a) a $t$ in your final answer (hint: it disappears when you integrate over it) and (b) getting things divided by 0 and/or $t$.

$\int_0^x \cos (\pi t) dt = \sin (\pi x) / \pi$, which takes values from $-1/\pi$ to $1/\pi$ as $x$ varies. If we take the two-sided integral, then we have $\int_{-x}^x \cos (\pi t) dt = (\sin (\pi x) - \sin (-\pi x)) / \pi = 2\sin (\pi x) / \pi$, because the function we are integrating is even (and hence has the same area to the left and right of the y-axis). This also behaves sinusoidally as $x$ varies, so even if we just try to take the limit as $x \rightarrow \infty$ we can see we're not going to get a nice limit.

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  • $\begingroup$ In the context of distributions $\int_{-\infty}^\infty \cos(k x)\,dx=2\pi \delta(k)$. So, in distribution, $$\int_{-\infty}^\infty \cos(\pi x)\,dx=0$$ $\endgroup$ – Mark Viola Jan 17 at 20:14

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