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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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    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