2
$\begingroup$

This is a question I came across and I cannot find the answer.

By using a substitution involving the tangent function, show that $$\int_0^1\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx=\frac{\pi}{4}$$

My attempt

I use trig substitution, by saying $$\tan(\theta)=\frac{y}{x}$$ which means $$x\sec^2(\theta)\,d\theta=dy$$ Also, it should be noted that because of this $$x\sec(\theta)=\sqrt{x^2+y^2}$$ $$x^4\sec^4(\theta)=(x^2+y^2)^2 $$ Thus, when I substitute this information into the integral, I get $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2-(x^2\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)} \, d\theta \,dx$$ Then, this simplifies to $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx$$ $$\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2\sec^2(\theta)}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx=\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})}{\frac{1}{x}}\,d\theta \,dx$$ which leads to $$\int_0^1 \left[\frac \theta x \right]_0^{\arctan\left(\frac{1}{x}\right)} \,dx = \int_0^1 \frac{\arctan\left(\frac{1}{x}\right)}{x} \, dx_{(3)} $$ At this point I am stuck. How do I evaluate this integral. Am I on the right path? Wolfram Alpha gives an answer other than $\frac{\pi}{4}$ for (3), so I am not sure where I am wrong.

$\endgroup$
1
  • $\begingroup$ Perhaps this is your culprit: $1 - \tan^2\theta \ne \sec^2\theta$. $\endgroup$ – suneater May 12 '16 at 6:17
3
$\begingroup$

Let us start considering $$I=\int\frac{x^2-y^2}{(x^2+y^2)^2}dy$$ Defining $$y=x\tan(\theta)\implies dy=x \sec ^2(\theta )\implies I=\int \frac{\cos (2 \theta )}{x}\,d\theta=\frac{\sin (2 \theta )}{2 x}$$ Back to $x$ $$I=\frac{y}{x^2+y^2}$$ So, $$\int_0^1\frac{x^2-y^2}{(x^2+y^2)^2}dy=\frac 1 {1+x^2}$$ So, you are left with $$\int_0^1\frac {dx} {1+x^2}$$ Repeat the same change of variable.

$\endgroup$
2
  • $\begingroup$ how do you know that those substitutions become $$\frac{cos(2\theta)}{x}$$ $\endgroup$ – user278039 May 12 '16 at 5:28
  • 1
    $\begingroup$ @user278039. Classical trigonometric identities and simplifications. $\endgroup$ – Claude Leibovici May 12 '16 at 5:55
3
$\begingroup$

Starting with your work... $$\begin{align} \int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{x^2(1-\tan^2(\theta))}{x^4\sec^4(\theta)}{x\sec^2(\theta)}\,d\theta \,dx =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{(1-\tan^2(\theta))}{x\sec^2(\theta)}\,d\theta \,dx \\\\ =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{(\cos^2\theta-\sin^2\theta)}{x}\,d\theta \,dx \\\\ =&\int_0^1\int_{\theta=0}^{\arctan(\frac{1}{x})} \frac{\cos2\theta}{x}\,d\theta \,dx \\\\ =&\frac12\int_0^1 \frac{\sin2\theta}{x}\bigg|_0^{\arctan(1/x)} \,dx \\\\ = &\int_0^1 \frac{\sin\theta\cos\theta}{x}\bigg|_0^{\arctan(1/x)} \,dx \\\\ =&\int_0^1 \frac{x}{x(1+x^2)} \,dx = \arctan(1)=\pi/4 \end{align}$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.