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This is my first question on this site, so please pardon any nuanced formatting errors.

My friend and I were discussing the following integral yesterday:

$$\int_0^3 \frac{1}{\sqrt{x-3}} \, \mathrm{d}x$$

I discovered that the integral evaluates to $-2i\sqrt{3}$ This confuses me. When I studied calculus 1, I was taught that the definite integral can be thought of as a way to calculate the "area under the curve" or the accumulation of a value. How can an area have a complex value?

Thus, I know that my conceptual understanding of integrals is flawed. How can I reconcile the concept of "area" with a complex value? Or, what is a "better" (for lack of a better word) way to conceptualize the integral?

Thanks in advance!

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    $\begingroup$ Your function has complex values for $x\lt 3$ $\endgroup$ – Riccardo.Alestra Jul 20 '16 at 17:09
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    $\begingroup$ There is no curve for $x \le 3$ in your function. No curve no area. $\endgroup$ – Pentapolis Jul 20 '16 at 21:26
  • $\begingroup$ The function $\frac {1}{\sqrt{x-3}}$ has only 0 and purely imaginary values for $0\le x \le 3$ so the integral will be purely imaginary. The curve exists only in "imaginary" plane so the area under it will only be in the imaginary plane. ... Or another way to think of this is $\int_{0}^3 \frac 1{\sqrt{x - 3}} dx = \int_{0}^3 \frac 1{i\sqrt{3 -x}} dx = \frac 1i \int_{0}^3 \frac 1{\sqrt{x-3}} dx$. Does that make it clearer. $\endgroup$ – fleablood Jul 20 '16 at 22:47
  • $\begingroup$ As a hint. . Try graphing $f(x) = \frac 1{\sqrt{x-3}}$ for $0 \le x \le 3$. You'll notice it has no real value. The graph will not be graphed on the xy plane $x,y \in \mathbb R$. but on the xy plane $x \in \mathbb R; y \in i\mathbb R$. $\endgroup$ – fleablood Jul 20 '16 at 23:03
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The truth is the concept in your mind (i.e., geometric area and integration) applies to real scalar functions with single variable, while you're not integrating a real function.

In your case, assuming the integrand function ($1/\sqrt{x-3}$) to be a real scalar function, it is not defined over the integration interval $(0,3)$. But, if you assume it to be a complex function like $f(z)=1/\sqrt{z-3}$, then you can integrate it over that interval, and of course, you will get an imaginary result since the function has pure imaginary values throughout the integration interval. Obviously, if you change the integrand function to $g(x)=1/\sqrt{3-x}$, then the result would be a real number equivalent to the geometric area you're looking for.

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The area under the curve idea applies to real-valued functions. This function is imaginary for almost all values of the area of integration.

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Your problem is 'well defined' once you specifies the $\ds{\root{z - 3}}$ branch-cut. For instance, $$ \root{z - 3} = \root{\verts{z - 3}}\exp\pars{{\,\mathrm{arg}\pars{z - 3} \over 2}\,\ic} \,,\quad z \not= 3\,,\quad 0 < \,\mathrm{arg}\pars{z - 3} < 2\pi $$ and the branch cut is along the 'line segment' $$ \braces{z = 3 + r\expo{\ic\alpha}\,,\quad r \geq 0\,,\quad 0 \leq \alpha <\pi\quad \mbox{or}\quad \pi < \alpha < 2\pi} $$


In that case, \begin{align} &\color{#f00}{% \int_{0}^{3}{\dd x \over \root{\verts{x - 3}}\expo{\ic\pars{\pi - \alpha}}/2}} = -\ic\expo{\ic\alpha/2}\int_{0}^{3}{\dd x \over \root{3 - x}} = \color{#f00}{-2\root{3}\expo{\ic\alpha/2}\,\,\ic} \end{align}

The result $\ds{-2\root{3}\,\ic}$ is given for a branch-cut along the 'positive real axis'. Namely,$\ds{\braces{z = x + 0\ic\,,\quad x \geq 3}}$ which is equivalent to $\ds{\alpha = 0}$.

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Notice for $0 \le x \le 3$ then $x - 3 \le 0$ so $\sqrt{x - 3} = \sqrt{-1*(3 - x)} = i \sqrt{3 - x}$ is strictly $0$ or purely imaginary.

So $\int_{0}^3 \frac 1{\sqrt{x - 3}} dx = \int_{0}^3 \frac 1{i\sqrt{3-x}} dx = \frac 1i \int_{0}^3 \frac 1{\sqrt{3-x}} dx$.

Now $\int_{0}^3 \frac 1{\sqrt{3-x}} dx= - \sqrt{3-x}|_{0}^3= \sqrt{3} $ is purely real, $\frac 1i $ (which equals $-i$ BTW) is purely imaginary. so $\int_{0}^3 \frac 1{\sqrt{x - 3}} dx = \int_{0}^3 \frac 1{i\sqrt{3-x}} dx = \frac 1i \int_{0}^3 \frac 1{\sqrt{3-x}} dx = -i\sqrt{3}$ is purely imaginary.

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