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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 $0<x\le b$.

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?

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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. –  Ross Millikan Sep 29 '11 at 0:43
1  
Why do you think it should have a logarithm? Did you manage to derive a solution that has a logarithm in it? –  J. M. Sep 29 '11 at 1:10
    
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. –  liberias Sep 29 '11 at 1:12
    
Note inverse trigonometric functions have logarithmic forms. –  anon Sep 29 '11 at 1:12
2  
Can you include that expression in your question? –  J. M. Sep 29 '11 at 1:28
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