We cannot take the definite integral of $\tan(x)$ on the interval $[0, \pi/2]$, just like we can't take the definite integral of $1/x$ on $]0, 1]$; but what would happen if we took the definite integral of $\tan(x)$ between $[0, \infty[$ ? Would it converge and lead to a result ? just like: $$\int_0^{\infty} e^{-2x} \,dx$$
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Is $tg(x)$ the tangent function? – davidlowryduda Apr 24 '16 at 14:46
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yes; that's the way we write it in Hungary. – MattMatt Apr 24 '16 at 14:48
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1What would happen is that one would get an "even more" undefined quantity. – Did Apr 24 '16 at 14:51
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1Yeah this isn't Riemann integrable over the interval. Infinities don't cancel, though it's tempting. – KR136 Apr 24 '16 at 14:53
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hmm... interesting; according to my plotting program, the area is around -1.043E+7 – MattMatt Apr 24 '16 at 14:53
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@Mattmatt Plotting programs do not handle "cancelling" infinities very well, i.e. they sometimes believe that they actually cancel (since their methods are unable to see any problem with the integration in the first place). I guess it would say that $\int_{-1}^1\frac1xdx$ is well-defined as well. – Arthur Apr 24 '16 at 14:56
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@Arthur nope; for 1/x over ]0, 1], it says undefined – MattMatt Apr 24 '16 at 14:57
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it's actually really reliable – MattMatt Apr 24 '16 at 14:58
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@MattMatt "aroud" minus ten millions... The point is that, depending on the subdivisions of $[0,\infty[,\setminus \left{\left(k+\frac12\right)\pi,:,k\in\Bbb N\right}$ your algorithm considers, you'll get completely different results (that's somehow what "not Riemann integrable" means). – Apr 24 '16 at 14:58
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that makes more sense, yes – MattMatt Apr 24 '16 at 15:00
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@MattMatt Arthur mentioned the integral of $1/x$ on $(-1,1)$, not on $(0,1)$ hence the "undefined" you got for the latter is offtopic for the former. – Did Apr 24 '16 at 15:35
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That integral move between ∞ and 0. – Takahiro Waki Apr 24 '16 at 16:15
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It is equal to $\ln 2$ – Anixx Mar 15 '20 at 21:31
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In order for the definite integral on $]0,\infty[$ to be well-defined, the integrand must approach 0 as x approaches $\infty$. Clearly, tan(x) doesn't do that, so we can't talk about $\int_0^\infty tan(x)dx$.
What about $\int_0^{\pi/2} \tan(x)dx$?
Well since the indefinite integral of $tan(x)$ is $\ln[\sec(x)]$, we see that $\lim_{x\to \pi/2^-}[\ln \sec(x)] = \ln \infty = \infty$ is unbounded, and thus the area of even one period of tangent does not converge.
Ben G.
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