MHB Where Do I Go From Here in Evaluating This Integral?

[Pull and Twist]

I'm trying to evaluate the following problem...
$$\int_{0}^{\infty} \frac{x\arctan\left({x}\right)}{(x^2+1)^2}\,dx$$

$$x=\tan\left({\theta}\right)$$
$$\theta=\arctan\left({x}\right)$$
$$dx=\sec^2\left({\theta}\right)d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\tan\left({\theta}\right)\sec^2\left({\theta}\right)}{\sec^4\left({\theta}\right)}\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\tan\left({\theta}\right)}{\sec^2\left({\theta}\right)}\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \left(\frac{\theta}{\sec\left({\theta}\right)}\right)\left(\frac{\tan\left({\theta}\right)}{\sec\left({\theta}\right)}\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \left(\frac{\theta}{\sec\left({\theta}\right)}\right)\left(\sin\left({\theta}\right)\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \theta\cos\left({\theta}\right)\sin\left({\theta}\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta2\cos\left({\theta}\right)\sin\left({\theta}\right)\,d\theta}{2}$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\sin\left({2\theta}\right)\,d\theta}{2}$$

$$\lim_{{t}\to{\infty}}\frac{1}{2}\int_{0}^{t} \theta\sin\left({2\theta}\right)\,d\theta$$

Integration by Parts
$$u=\theta$$
$$du=d\theta$$
$$dv=\sin\left({2\theta}\right)d\theta$$
$$v=-\frac{\cos\left({2\theta}\right)}{2}$$$$\lim_{{t}\to{\infty}}\frac{1}{2}\left\{\left[-\frac{\theta\cos\left({2\theta}\right)}{2}\right]+\int_{0}^{t}\cos\left({2\theta}\right)\,d\theta\right\}$$

$$\lim_{{t}\to{\infty}}\frac{1}{2}\left\{\left[-\frac{\theta\cos\left({2\theta}\right)}{2}\right]+\frac{\sin\left({2\theta}\right)}{4}\right\}$$

Where do I go from here??

 

[Greg]

Hint:

$$\int^{\frac{\pi}{2}}_0\frac{\theta\tan\theta\sec^2\theta}{\sec^4\theta}\,\text{d}\theta=\frac{\pi}{8}$$

 

[MarkFL]

I would begin by using IBP straightaway:

$$u=\arctan(x)\,\therefore\,du=\frac{1}{x^2+1}\,dx$$

$$dv=\frac{x}{\left(x^2+1\right)^2}\,\therefore\,v=-\frac{1}{2\left(x^2+1\right)}$$

And so we have:

$$\int_0^{\infty}\frac{x\arctan(x)}{\left(x^2+1\right)^2}\,dx=-\frac{1}{2}\left[\frac{\arctan(x)}{x^2+1}\right]_0^{\infty}+\frac{1}{2}\int_0^{\infty}\frac{1}{\left(x^2+1\right)^2}\,dx$$

Now since:

$$\lim_{x\to\infty}\frac{\arctan(x)}{x^2+1}=\lim_{x\to0}\frac{\arctan(x)}{x^2+1}=0$$, we have:

$$\int_0^{\infty}\frac{x\arctan(x)}{\left(x^2+1\right)^2}\,dx=\frac{1}{2}\int_0^{\infty}\frac{1}{\left(x^2+1\right)^2}\,dx$$

Rewriting the integrand on the right, we have:

$$\int_0^{\infty}\frac{x\arctan(x)}{\left(x^2+1\right)^2}\,dx=\frac{1}{4}\int_0^{\infty}\frac{\left(x^2+1\right)-2x^2+\left(x^2+1\right)}{\left(x^2+1\right)^2}\,dx$$

Making the observation that:

$$\frac{d}{dx}\left(\frac{x}{x^2+1}\right)=\frac{\left(x^2+1\right)-2x^2}{\left(x^2+1\right)^2}$$

$$\frac{d}{dx}\left(\arctan(x)\right)=\frac{1}{x^2+1}=\frac{x^2+1}{\left(x^2+1\right)^2}$$

Can you proceed?

 

[Euge]

I'm trying to evaluate the following problem...
$$\int_{0}^{\infty} \frac{x\arctan\left({x}\right)}{(x^2+1)^2}\,dx$$

$$x=\tan\left({\theta}\right)$$
$$\theta=\arctan\left({x}\right)$$
$$dx=\sec^2\left({\theta}\right)d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\tan\left({\theta}\right)\sec^2\left({\theta}\right)}{\sec^4\left({\theta}\right)}\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\tan\left({\theta}\right)}{\sec^2\left({\theta}\right)}\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \left(\frac{\theta}{\sec\left({\theta}\right)}\right)\left(\frac{\tan\left({\theta}\right)}{\sec\left({\theta}\right)}\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \left(\frac{\theta}{\sec\left({\theta}\right)}\right)\left(\sin\left({\theta}\right)\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \theta\cos\left({\theta}\right)\sin\left({\theta}\right)\,d\theta$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta2\cos\left({\theta}\right)\sin\left({\theta}\right)\,d\theta}{2}$$

$$\lim_{{t}\to{\infty}}\int_{0}^{t} \frac{\theta\sin\left({2\theta}\right)\,d\theta}{2}$$

$$\lim_{{t}\to{\infty}}\frac{1}{2}\int_{0}^{t} \theta\sin\left({2\theta}\right)\,d\theta$$

Integration by Parts
$$u=\theta$$
$$du=d\theta$$
$$dv=\sin\left({2\theta}\right)d\theta$$
$$v=-\frac{\cos\left({2\theta}\right)}{2}$$$$\lim_{{t}\to{\infty}}\frac{1}{2}\left\{\left[-\frac{\theta\cos\left({2\theta}\right)}{2}\right]+\int_{0}^{t}\cos\left({2\theta}\right)\,d\theta\right\}$$

$$\lim_{{t}\to{\infty}}\frac{1}{2}\left\{\left[-\frac{\theta\cos\left({2\theta}\right)}{2}\right]+\frac{\sin\left({2\theta}\right)}{4}\right\}$$

Where do I go from here??

Hi PullandTwist,

I think you're having trouble because either your upper limits of integration are off or your're taking the wrong limit. After performing the trig substitution $x = \tan(\theta)$, the upper limits of your integrals should be $\tan^{-1}(t)$ instead of $t$; otherwise, your limit should be as $t\to \pi/2^{-}$, not $t\to \infty$.