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A problem in our calculus text requires the evaluation of $\displaystyle\int_0^1\frac{\sqrt{1-y^2}}{1+y^2}dy$,

and I have evaluated it by substituting $y=\sin\theta$ (or $y=\tanh u$) and then using another substitution and partial fractions; but I would like to find out if there is a simpler way to find this integral that does not involve trig substitution (or hyperbolic substitution).

(I know some MSE people dislike this type of question, so I apologize in advance.)

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    $\begingroup$ you can find it by Taylor series $\endgroup$ – E.H.E Jun 8 '15 at 20:30
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    $\begingroup$ @Essam, sounds pretty gruesome. $\endgroup$ – Simon S Jun 8 '15 at 20:35
  • $\begingroup$ You can solve this using the method of contour integration, but this is maybe bnot what you want $\endgroup$ – tired Jun 8 '15 at 21:04
  • $\begingroup$ Another way of doing this is the so called euler substituion. But i'm not sure if you will be really faster. Nevertheless, it will work out definitly $\endgroup$ – tired Jun 8 '15 at 21:22
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    $\begingroup$ You can also use hyperbolic trig substitution or Euler substitution. I once wrote a post evaluating one integral with all three techniques. They all work in high generality. $\endgroup$ – davidlowryduda Jun 9 '15 at 19:44
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Hint:

$$t=y^2$$

$$dy=\frac{1}{2\sqrt t}dt$$

then

$$I=\frac{1}{2}\int_0^1\frac{\sqrt{\frac{1-t}{t}}}{1+t}dt$$ $$\frac{1-t}{t}=u^2$$

so$$t=\frac{1}{1+u^2} $$

$$I=\int_0^{\infty}\frac{u^2}{(u^2+1)(u^2+2)}du=\int_0^{\infty}(\frac{2}{u^2+2}-\frac{1}{u^2+1})du$$ then you can solve it without trigonometric substitution by knowing that

$$\int_0^{\infty}\frac{1}{u^2+1}du=\frac{\pi}{2}$$

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    $\begingroup$ very neat! (+1) $\endgroup$ – tired Jun 9 '15 at 20:56
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Let we take the "gruesome" path. $$\begin{eqnarray*}\int_{0}^{1}\frac{\sqrt{1-x^2}}{1+x^2}\,dx&=&\sum_{n\geq 0}(-1)^n\int_{0}^{1}x^{2n}\sqrt{1-x^2}\,dx = \frac{\sqrt{\pi}}{4}\sum_{n\geq 0}(-1)^n\frac{\Gamma\left(n+\frac{1}{2}\right)}{\Gamma(n+2)}\\&=&\frac{\pi}{4}\sum_{n\geq 0}\frac{(-1)^n}{4^n(n+1)}\binom{2n}{n}=\left.\frac{\pi}{2}\cdot\frac{1-\sqrt{1-x}}{x}\right|_{x=-1}\\&=&\color{red}{\frac{\pi}{2}(\sqrt{2}-1)}.\end{eqnarray*}$$ Not so painful, after all, if one recalls the generating function of the Catalan numbers.

Steps involved:

  • expansion of $\frac{1}{1+x^2}$ as a geometric series;
  • integration of $x^{\alpha}(1-x)^{\beta}$ over $[0,1]$ through the Euler beta function;
  • rewriting in terms of central binomial coefficients;
  • evaluation through the generating function for the Catalan numbers.
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  • $\begingroup$ Thanks for your answer, which doesn't look so gruesome after all. If you have time, though, would you please explain your 2nd and 3rd equalities (or provide a link that would help me understand them). $\endgroup$ – user84413 Jun 10 '15 at 18:05
  • $\begingroup$ @user84413: I added a brief explanation of the steps involved. $\endgroup$ – Jack D'Aurizio Jun 10 '15 at 18:27
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I will try to give an approach to this integral using complex analysis:

First of all we would like to have a form of the integral which is most easily tractable by contour integration, which (at least for me) means that

  • it is as obivious as possible to guess a contour which can be used

  • the pole/cut structure is as clear as possible

One way of doing so, is exploiting the parity of the integrand and transform $y\rightarrow 1/x$. We get: $$ I=\frac{1}{2}P\int_{-1}^{1}\frac{1}{x}\frac{\sqrt{x^2-1}}{1+x^2}dx $$

Here $P$ denotes Cauchy's principal value. We now may consider the complex function

$$ f(z)=\frac{1}{z}\frac{\sqrt{z^2-1}}{1+z^2} $$

Choosing the standard branch of logarithm, we have a cut on the interval $[-1,1]$, furthermore we have singularities at $\{\pm i,0\}$ where the first two are harmless but the one at zero is on the cut and will need to be handeled with care.

Now may choose a contour which encloses the branch cut, and avoids the singularity at 0. we get:

$$ \oint f(z)dz = \underbrace{\int_{-1}^{-\epsilon}f(x_+)+\int_{\epsilon}^{1}f(x_+)}_{2I}+\int_{\text{arg}(z)\in(\pi,0], |z|=\epsilon}f(z)dz\\-\underbrace{\int_{-1}^{-\epsilon}f(x_-)-\int_{\epsilon}^{1}f(x_-)}_{-2I}-\int_{\text{arg}(z)\in[0,-\pi), |z|=\epsilon}f(z)dz=\\ 4I+\underbrace{2\int_{\text{arg}(z)\in(\pi,0], |z|=\epsilon}f(z)dz}_{2 \times \pi i\ \times \text{res}[f(z),z=0] }=\\ 4I +2\pi \quad (1) $$

Where $\text{res}[f(z),z=0]=i$ because we calculated the residue above the branch cut. Furthermore $f(x_{\pm})$ denotes about which side of the cut we are talking: $\pm$ above/below. Also the limit $\epsilon \rightarrow 0$ is implicit.

Now comes the trick: By looking at the exterior of the contour we can also write (please note that we now enclose the singularities in opposite direction compared to above)

$$ \oint f(z)dz=-2\pi i \times(\text{res}[f(z),z=i]+ \text{res}[f(z),z=-i])=2\sqrt{2}\pi \quad (2) $$

Equating $(1)=(2)$

$$ 4I+2\pi=2\sqrt{2}\pi\\ $$ or $$ I=\frac{\pi}{2}\left(\sqrt{2}-1\right) $$

which is the same result as the one obtained by trig. substitution.

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  • $\begingroup$ Thanks for providing this answer. (There may be a sign error, though.) $\endgroup$ – user84413 Jun 9 '15 at 17:47
  • $\begingroup$ thanks for pointing this out. i will a have a look at it later! $\endgroup$ – tired Jun 9 '15 at 18:41

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