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Let C be a rectangle with corners $\pm 2\pm 3i$ in the anticlockwise direction. Find $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$

I tried:

Split the contour in to 4 parts: $C_1,C_2,C_3,C_4$.

Let's say $C_1$ is the the bottom side of the rectangle. So,

$C_1: f(t) =t-3i, -2\le t\le 2$

Then I find $$\int_{-2}^{2}\frac{1}{t-3i}+\frac{1}{t-3i-1}dt$$

Similarly, I would find the integrals over $C_2, C_3$ and $C_4$. Then, the final answer is the sum of $C_1,C_2,C_3,C_4$.

Is this correct?

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    It's much simpler to use the Residue Theorem. – 2017-02-25
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    Your first integral has a lower limit of $c$ and no upper limit – 2017-02-25
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    @mrnovice that is standard notation... – 2017-02-25
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    Oh ok, this is a topic beyond me then – 2017-02-25
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    The two poles are inside the contour, just use the residue theorem. – 2017-02-25
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    Or Cauchy's integral formula. – 2017-02-25
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    Anyways, to the actual question, yes, you can and probably should do what you are doing. – 2017-02-25
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    It will be so painful to consider the integral as is because you have to consider the branch of logarithms. – 2017-02-25
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    @SimplyBeautifulArt. I haven't learned the Residue Theorem, yet, so I can't use that. But how do I use the Cauchy Integral Theorem if f(z) is not analytic at z = 0 and z = 1 and 0, 1 are both inside C? Or am I wrong? – 2017-02-25

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From the comments, it appears that the OP is receptive to seeing how Cauchy's Integral Formula can be used to evaluate the integral of interest.

Let $C$ be a closed rectifiable contour with winding number $1$ about a point $z_0\in \mathbb{C}$. Cauchy's Integral Formula states that if $f$ is analytic on the open region enclosed by $C$ and continuous on $C, then

$$f(z_0)=\frac{1}{2\pi i}\oint_C \frac{f(z)}{z-z_0}\,dz \tag 1$$

In $(1)$, set $f(z)=1$, $z_0=0$ and $C$ to be the rectangle defined in the OP. Then from $(1)$ we have

$$1=\frac{1}{2\pi i}\oint_C \frac{1}{z}\,dz$$

whereupon solving for $\oint_C \frac{1}{z}\,dz$ reveals

$$\oint_C \frac{1}{z}\,dz=2\pi i \tag 2$$

Similarly, set $f(z)=1$ and $z_0=1$. Then from $(1)$ we have

$$1=\frac{1}{2\pi i}\oint_C \frac{1}{z-1}\,dz$$

whereupon solving for $\oint_C \frac{1}{z}\,dz$ reveals

$$\oint_C \frac{1}{z-1}\,dz=2\pi i \tag3$$

Putting together $(2)$ and $(3)$ yields

$$\oint_C \left(\frac1z+\frac1{z-1}\right)\,dz=4\pi i$$

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    The integral I'm looking for is $$\int_c\frac{1}{z}+\frac{1}{z-1}dz$$, so shouldn't the correct answer be $4\pi i$? Also, do the corners of the rectangle not matter at all when using the cauchy integral formula? – 2017-02-25
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    Yes, you're correct. It's $4\pi i$. The corners don't matter. The contour need only be rectilinear. – 2017-02-25
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    @sucksatmath Since you're new to the site, I wanted to you know that after you have enough reputation points you can up vote answers too. -Mark – 2017-03-17