Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am currently trying to understand and implement a method from a paper. Inside the method there is an integral I have to solve by hand. I got the result in form of a series expansion but I don't understand the way to get to this result.

The integral is

\begin{align} \int\limits\left[\left(t-a_1\right)\cdot\left(t-\bar{a}_1\right)\cdot\left(t+a_1\right)\cdot\left(t-\bar{a}_1\right)\right]^{\frac{1}{2}}\frac{dt}{t^2} \end{align}

with $a_1=e^{k\pi i}$ and $\left\{k\in\mathbb{R}|0\le k\le\frac{1}{2}\right\}$ and the complex conjugate $\bar{a}_1$. The paper says the result is a serial expansion in the form of

\begin{align} \int\limits\left[1-\frac{1}{2}\left(a_1^2+\bar{a}_1^2\right)\frac{1}{t^2}-\frac{1}{8}\left(a_1^2-\bar{a}_1^2\right)^2\frac{1}{t^4}+\ldots\right]dt \end{align}

It looks as if the expression in the first formula can be transformed to something similar to the binomial series $\left(1-\frac{1}{x^2}\right)^{\frac{1}{2}}$ but right now I have no idea how. I tried to several approaches, for example

\begin{align} &\int\limits\left[\left(t^2-a_1^2\right)\cdot\left(t^2-\bar{a}_1^2\right)\right]^{\frac{1}{2}}\frac{dt}{t^2}\\ =&\int\limits\left[\left(t^4-\left(a_1^2+\bar{a}_1^2\right)\cdot t^2+a_1^2\cdot\bar{a}_1^2\right)\right]^{\frac{1}{2}}\frac{dt}{t^2}\\ =&\int\limits\left[\left(1-\frac{\left(a_1^2+\bar{a}_1^2\right)}{t^2}+\frac{a_1^2\cdot\bar{a}_1^2}{t^4}\right)\right]^{\frac{1}{2}}dt \end{align}

but I still can't find a way to obtain the series expression. Even if $a_1\cdot\bar{a}_1=1$ I still have

\begin{align} &\int\limits\left[\left(1-\frac{\left(a_1^2+\bar{a}_1^2\right)}{t^2}+\frac{1}{t^4}\right)\right]^{\frac{1}{2}}dt \end{align}

In my opinion I got one term too much in the integrand. Is there a way to get rid of the term with $t^4$?

Could someone please show me a way? What am I missing?

The integral and series expansion can be found in MilneThomson on page 153.

share|cite|improve this question
hint: $(a+b)(a-b)=a^2-b^2$ – oldrinb Sep 4 '13 at 17:45
Thanks, I tried that, but from $\int\limits\left[\left(t^2-a_1^2\right)\cdot\left(t^2-\bar{a}_1^2\right)\right]‌​^{\frac{1}{2}}\frac{dt}{t^2}$ I still have no idea how to get to something in the form of $(1-x)^{\frac{1}{2}}$ of the integrand. – user93130 Sep 4 '13 at 20:12
You can edit your original question if you are still having difficulty. – Emily Sep 4 '13 at 21:46
Sorry to interrupt but is $a_1=\pm1$? – Did Sep 5 '13 at 7:07
Well, $a_1=e^{k\pi i}$, so $|a_1|=1$ – Martin Sep 5 '13 at 7:17

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.