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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?


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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). – Henning Makholm Nov 27 '11 at 2:32
up vote 15 down vote accepted

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 and here; this is not really a good definition of $i$. – Zev Chonoles Nov 28 '11 at 7:16
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. – dku.rajkumar Nov 28 '11 at 9:04
@ZevChonoles I don't see the problem (there is completely general definition $R(\sqrt d):=R[x]/(x^2-d)$, if you wish) – Grigory M Nov 28 '11 at 9:08
@Grigory: That definition is great; but what I would not say in that situation is $x=\sqrt{d}$, or $x=d^{1/2}$. – Zev Chonoles Nov 28 '11 at 9:34
@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 $i=(-1)^{1/2}$. – Zev Chonoles Nov 28 '11 at 9:39

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