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How to calculate $\int_{-1}^1 \frac{dt}{\sqrt{1-t^2}(t-x)}$ for $|x|>1$ by Residue theorem? I could do is just as: $$I(x)=\int_{-\pi/2}^{\pi/2}\frac{d\theta}{\sin\theta-x}\\ =\frac{1}{2}\int_{\pi}^\pi \frac{d\tau}{\sqrt{\frac{1-\cos \tau}{2}}-x}.$$

It is difficult to do now by replacing $z=e^{i\tau}$.

XLDD
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3 Answers3

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You shouldn't have done that last step - as you have already realised, it doesn't help.

Instead you should notice that $$\int_0^{\pi/2} \frac{d\theta}{\sin\theta-x} =\int_{\pi/2}^\pi \frac{d\theta}{\sin\theta-x}$$ - to prove this either think of the graph of the integrand, or substitute $\theta\gets\pi-\theta$. Doing the same for the integral from $-\pi/2$ to $0$, your integral becomes $$I(x)=\frac{1}{2}\int_{-\pi}^\pi \frac{d\theta}{\sin\theta-x}\ ,$$ and now you will find that the usual complex substitution works fine.

David
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  • The integral in the RHS should have $-\cos\left(\theta\right)$ instead of $\sin\left(\theta\right)$. – Felix Marin Mar 11 '14 at 07:25
  • @FelixMarin, no it should not: $\sin(\pi-\theta)=\sin\theta$. – David Mar 11 '14 at 09:59
  • $\displaystyle{\large\int_{0}^{\pi/2}{{\rm d}\theta \over \sin\left(\theta\right) - x} =\int_{0 + \pi/2}^{\pi/2 + \pi/2}{{\rm d}\theta \over \sin\left(\theta - \pi/2\right) - x}=\int_{\pi/2}^{\pi}{{\rm d}\theta \over -\cos\left(\theta\right) - x}}$. – Felix Marin Mar 11 '14 at 10:03
  • But you are substituting $\theta+\pi/2$ and I am not. Your answer appears to be a possible alternative but mine is also correct. Moreover, in yours the integrand has changed and therefore, as far as I can see, cannot easily be combined with the original integral. – David Mar 11 '14 at 10:05
  • You are right. Both results are equivalent. I used the one in my last comment to perform the calculation. Sorry. Thanks. – Felix Marin Mar 11 '14 at 10:13
  • Not a problem! I didn't actually complete the evaluation in my answer because I judged from the question that the OP is familiar with the $z=e^{i\theta}$ substitution. – David Mar 11 '14 at 10:17
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I can provide you with a "raw" answer while I figure it out. http://symbolab.com/solver/step_by_step/%5Cint_%7B-1%7D%5E%7B1%7D(%5Cfrac%7B1%7D%7B%5Csqrt%7B1-t%5E%7B2%7D%7D(t-x)%7D)dt%20

louie mcconnell
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$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle #1 \right\rangle} \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace} \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left( #1 \right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{{\rm I}\pars{x} = \int_{-\pi/2}^{\pi/2}{\dd\theta \over \sin\pars{\theta} - x}: \ {\large ?}\,,\qquad\verts{x} > 1}$

\begin{align} {\rm I}\pars{x}&=\int_{0}^{\pi}{\dd\theta \over -\cos\pars{\theta} - x} =\half\bracks{% \int_{0}^{\pi}{\dd\theta \over -\cos\pars{\theta} - x} + \int_{0}^{\pi}{\dd\theta \over \cos\pars{\theta} - x}} \\[3mm]&=\half\int_{0}^{\pi}{-2x \over x^{2} - \cos^{2}\pars{\theta}}\,\dd\theta =-x\int_{0}^{\pi}{1 \over x^{2} - \bracks{1 + \cos\pars{2\theta}}/2}\,\dd\theta \\[3mm]&=-x\int_{0}^{2\pi}{\dd\theta \over 2x^{2} - 1 - \cos\pars{\theta}} =-x\int_{\verts{z} = 1} {\dd z/\pars{\ic z} \over 2x^{2} - 1 - \pars{z^{2} + 1}/\pars{2z}} \\[3mm]&=-2x\ic\int_{\verts{z} = 1} {\dd z \over z^{2} - 2\mu z + 1}\,,\qquad\qquad\mu \equiv 2x^{2} - 1 > 1 \tag{1} \end{align}

The zeros of $z^{2} - 2\mu z + 1 = 0$ are given by: \begin{align} {\rm z_{-}}\pars{\mu} &= \mu - \root{\mu^{2} - 1}={1 \over \mu + \root{\mu^{2} - 1}} < 1 \\ {\rm z_{+}}\pars{\mu} &= \mu + \root{\mu^{2} - 1} > 1 \end{align}

By replacing these results in $\pars{1}$, we find: \begin{align} \color{#00f}{\large{\rm I}\pars{x}}&=-2xi\pars{2\pi\ic}\, {1 \over 2{\rm z_{-}}\pars{\mu} - 2\mu} ={2\pi x \over -\root{\mu^{2} - 1}} =-\,{2\pi x \over \root{\pars{\mu - 1}\pars{\mu + 1}}} \\[3mm]&=-\,{2\pi x \over \root{\pars{2x^{2} - 2}\pars{2x^{2}}}} =\color{#00f}{\large -\,{\pi\sgn\pars{x} \over \root{x^{2} - 1}}} \end{align}

Felix Marin
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