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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!

Michael Hardy
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badatmath
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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). – hmakholm left over Monica Nov 27 '11 at 02:32

4 Answers4

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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.

Michael Hardy
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You can treat $i$ as a constant:

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

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

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

Chris Taylor
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dku.rajkumar
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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 07:16
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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. – dku.rajkumar Nov 28 '11 at 09:04
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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) – Grigory M Nov 28 '11 at 09:08
  • @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 09:34
  • @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 09:39
  • @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. – dku.rajkumar Nov 28 '11 at 10:05
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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. – Grigory M Nov 28 '11 at 20:18
  • @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. – Zev Chonoles Nov 28 '11 at 20:39
  • 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 $$\mathbb{R}[x]/(x^2+1)\to\mathbb{R}[y]/(y^2+1),$$ – Zev Chonoles Nov 28 '11 at 20:54
  • 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. – Zev Chonoles Nov 28 '11 at 20:54