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How to evaluate this integral? $$\displaystyle I_1=\int\frac{\left(\frac{a}{y}-\frac{a}{b}\right)^{1/2}}{1-\frac{a}{y}}\mathrm {d}y$$ Here $a$ and $b$ are real constants and $y$ real variable.

It is solvable by a method of integration by substitution, where new variable $t$ is introduced by some trigonometric combination, e.g. $y=b\cos^2t$. However the solution is not as expected. Can someone suggest a better new variable or another method?

The solution obtained via $y=b\cos^2t$ is $$I_1=-\frac{1}{2}\sqrt{ab}\left\{2\left(1-\frac{a}{b}\right)\arccos\left(\frac{y}{b}\right)+4\sqrt{\frac{a}{b}\left(\frac{a}{b}-1\right)}\arctan\left(\frac{\frac{b}{y}-1}{1-\frac{b}{a}}\right)^{1/2}-2\sqrt{\frac{y}{b}\left(1-\frac{y}{b}\right)}\right\}$$

The solution should be composed of functions $\arctan$, $\log$ and square roots.



It turned out that ther integral $I_1$ is not exactly what I should solve. (It is related to integral $I_2$ by addition of a number of complicated terms.)

The following integral is the problem: $$I_2=-\left(1-\frac{a}{b}\right)^{1/2}\int_{b}^{x}\left(1-\frac{a}{y}\right)^{-1}\left(\frac{a}{y}-\frac{a}{b}\right)^{-1/2}dy\;,$$ where $0

Result from WolframAlpha is not returned.

The proposed solution has the form $$I_2=\left(\frac{b}{a}-1\right)^{1/2}b\sqrt{\frac{b}{x}-1}+a\left(\frac{b}{a}-1\right)^{1/2}\left(2+\frac{b}{a}\right)\arctan\left(\sqrt{\frac{b}{x}-1}\right)\sqrt{\frac{r_s}{x}}+a\log\left|\frac{\sqrt{\frac{b}{x}-1}+x^2}{\sqrt{\frac{b}{x}-1}-x^2}\right|$$

The question is whether this proposed solution is correct and how to derive it?

Again by introducing a new variable $y=b\cos^2t$, I came to this solution of indefinite integral in $I_2$: $$\left(-\left(1-\frac{a}{b}\right)^{1/2}\right)\left(-\frac{\sqrt{ab}}{2}\right)\left\{\left(2-\frac{a}{b}\right)\arccos\left(\frac{y}{b}\right)^{1/2}-2\left[\frac{y}{b}\left(\frac{y}{b}-1\right)\right]^{1/2}+4\left[\frac{a}{b}\left(\frac{a}{b}-1\right)\right]^{1/2}\arctan\left(\frac{\frac{b}{y}-1}{1-\frac{b}{a}}\right)^{1/2}\right\}$$

Here is the result of indefinite integral in $I_2$ from WolframAlpha.

Could this lead to the proposed solution which contains $\log$ and $\arctan$ with a different argument?

  • 3
    What solution do you get and why is it not as expected? Many indefinite integrals have alternate forms that are not obviously the same or differ only by a constant.2011-09-29
  • 1
    Why do you think it should have a logarithm? Did you manage to derive a solution that has a logarithm in it?2011-09-29
  • 0
    By introducing different new variable, there should be other functions present in the result. The question is, what are good choices of new variables in such situations, maybe $\cosh^2(t)$ ..., can't try them all.2011-09-29
  • 0
    Note inverse trigonometric functions have [logarithmic forms](http://en.wikipedia.org/wiki/Inverse_trigonometric_functions#Logarithmic_forms).2011-09-29
  • 0
    I have the solution of a definite integral over $y\in(b,x)$ and it contains a logarithm.2011-09-29
  • 2
    Can you include that expression in your question?2011-09-29
  • 0
    according to wolfram alpha solution is composed of log, square root and imaginary unit2011-09-29
  • 0
    Your edit has introduced a whole new integral. Please clarify which of the two integrals the question is about. The square root term was in the numerator in the original question; in the edit, it's in the denominator.2011-09-29
  • 0
    Yes, these are two different integrals, but $I_2$ is what I need to solve. They are very similar otherwise and maybe I should leave the first one for reference.2011-09-29
  • 0
    Well, have you tried differentiating both purported answers and see if they're the same?2011-09-29
  • 0
    Here is what I obtained in SWP $$\begin{eqnarray*} I_{1}&=&-2\sqrt{b-y}\sqrt{y}\left( -\frac{1}{2}\frac{\sqrt{a}}{\sqrt{b}}\right)\\ &&+\left( \arctan \frac{1}{2}\frac{-2y+b}{\sqrt{b-y}\sqrt{y}}\right) b\left( -\frac{1}{2}\frac{\sqrt{a}}{\sqrt{b}}\right)\\ &&-2\left( \arctan \frac{1}{2}\frac{-2y+b}{\sqrt{b-y}\sqrt{y}}\right) a\left( -\frac{1}{2}\frac{\sqrt{a}}{\sqrt{b}}\right)\\ &&+2\sqrt{a}\sqrt{-a+b}\ \text{arctanh}\frac{1}{2}\frac{ab+yb-2ay}{\sqrt{a} \sqrt{-a+b}\sqrt{b-y}\sqrt{y}}\left( -\frac{1}{2}\frac{\sqrt{a}}{\sqrt{b}} \right) \end{eqnarray*}$$2011-09-29
  • 0
    $$I_{1} =\int \frac{\left( \frac{a}{y}-\frac{a}{b}\right) ^{1/2}}{1-\frac{a}{y}}\mathrm{d}y$$2011-09-30

0 Answers 0