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How would you find, for instance, $\int_{0}^{4} i\> x dx$? Can you just treat $i$ as a constant, or do you have to do something more sophisticated?

Thanks!

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    As long as the _integration variable_ is real, you can just treat $i$ as a constant, as explained in the answers. However, if the variable can be _complex_, and entirely new vista of problems and possibilities opens up, and you shouldn't try to generalize your knowledge of real definite integrals to that setting without a course in complex analysis. (Not what you were asking about, just a warning).2011-11-27

4 Answers 4

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Yes nothing special. If $f$ and $g$ are real functions then $\int (f + i g) = \int f + i \int g$.

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Nothing special for situations like this, but if, for example, you're integrating $(1/x)\;dx$ not along the line from $0$ to $4$, but along a circle that winds once counterclockwise around $0$, then you may need something more sophisticated.

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You can treat $i$ as a constant:

$$\int_0^4 ix dx = i\int_0^4 xdx = i[x^2/2]_0^4 = i(8-0) = 8i$$

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"i" has one an only value , it never changes, hence it can be just taken out as constant.

$$\int i x \,dx = i\int x \,dx$$

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    -1: See the discussions [here](http://math.stackexchange.com/questions/63467/what-is-the-precise-definition-of-i) and [here](http://math.stackexchange.com/questions/84436/which-step-in-this-process-allows-me-to-erroneously-conclude-that-i-1); this is not really a good definition of $i$.2011-11-28
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    Thanks Man, I am well aware of the discussion done in the mentioned thread. The thing is I have not given any definition of "i" here.. I just tried to say that the value of i never changes hence it can not be a variable so "i" can be treated as constant.2011-11-28
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    @ZevChonoles I don't see the problem (there is completely general definition $R(\sqrt d):=R[x]/(x^2-d)$, if you wish)2011-11-28
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    @Grigory: That definition is great; but what I would not say in that situation is $x=\sqrt{d}$, or $x=d^{1/2}$.2011-11-28
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    @dku.rajkumar: If you don't want to call it a definition, fine; it is not really a good statement, then. I don't see how arguing that $i$ is a constant required that you claim that $i=(-1)^{1/2}$.2011-11-28
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    @Zev: (-1)^1/2 is a constant.. isn't it ?? and "i" can take only (-1)^1/2 value. So this way I tried to imply that "i" is a constant. may be its not 100% correct but that's what I meant.2011-11-28
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    @ZevChonoles Huh? Notation like $\mathbb Z[\sqrt{-5}]$ for rings and (say) $1+\sqrt{-5}$ for elements of such rings is absolutely standard.2011-11-28
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    @Grigory: Let $d$ be a positive integer. The ring $\mathbb{Z}[x]/(x^2-d)$ is **non-canonically** isomorphic to $\mathbb{Z}[\sqrt{d}]$ as we could equally well choose to send $x$ to either $\sqrt{d}$ or $-\sqrt{d}$. While it is a minor point, to be sure, I would usually avoid writing something like $x=\sqrt{d}$ in this situation, or would note that we are making a choice.2011-11-28
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    But even in the above, I was taking $d$ to be a positive integer, where we have an established - though of course still arbitrary - convention as to which element of $\mathbb{C}$ is specified by the expression "$\sqrt{d}$". Now, since neither $i$ nor $\sqrt{-1}$ really have any independent meaning, they are only really interpretable though this "abstract" square root process. I see the statement "$i=\sqrt{-1}$", or equivalently "$i=(-1)^{1/2}$", as basically equivalent to asserting that there is a canonical $\mathbb{R}$-algebra isomorphism $$\mathbb{R}[x]/(x^2+1)\to\mathbb{R}[y]/(y^2+1),$$2011-11-28
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    and it is the one sending $x$ to $y$, which is not correct, since the isomorphism sending $x$ to $-y$ is just as good. But dku.rajkumar has edited his answer (consequently, I have changed my downvote into an upvote), and I feel I have made something of a mountain out of a molehill, so I do not want to press the point further. Of course, if I have made any incorrect statements, I would welcome your correction.2011-11-28