Integration by Parts — Question 9

Question 9 – Solution

Let $$f(x)=x\arctan(x).$$ Both $x$ and $\arctan(x)$ are odd functions. Therefore, their product is even: $$\begin{align*} f(-x) &=(-x)\arctan(-x) \\ &=(-x)\bigl(-\arctan(x)\bigr) \\ &=x\arctan(x)=f(x). \end{align*}$$ Consequently, $$\int_{-1}^{1}x\arctan(x)\,dx =2\int_0^1x\arctan(x)\,dx. \tag{1}$$

Now use integration by parts on the integral over $[0,1]$. Choose $$u=\arctan(x),\quad dv=x\,dx, \qquad du=\frac{1}{1+x^2}\,dx,\quad v=\frac{x^2}{2}.$$ Then $$\begin{align*} \int_0^1x\arctan(x)\,dx &=\left.\frac{x^2}{2}\arctan(x)\right|_0^1 -\frac12\int_0^1\frac{x^2}{1+x^2}\,dx. \end{align*}$$ Multiplying by $2$ as in equation (1) gives $$\begin{align*} \int_{-1}^{1}x\arctan(x)\,dx &=\left.x^2\arctan(x)\right|_0^1 -\int_0^1\frac{x^2}{1+x^2}\,dx. \tag{2} \end{align*}$$

For the remaining integral, rewrite the integrand: $$\frac{x^2}{1+x^2} =\frac{(1+x^2)-1}{1+x^2} =1-\frac{1}{1+x^2}.$$ Therefore, $$\begin{align*} \int_0^1\frac{x^2}{1+x^2}\,dx &=\int_0^1\left(1-\frac{1}{1+x^2}\right)\,dx \\ &=\left[x-\arctan(x)\right]_0^1 \\ &=1-\frac{\pi}{4}. \end{align*}$$ The boundary term in equation (2) is $$\left.x^2\arctan(x)\right|_0^1 =1^2\arctan(1)-0^2\arctan(0) =\frac{\pi}{4}.$$ Substituting both results into equation (2), $$\begin{align*} \int_{-1}^{1}x\arctan(x)\,dx &=\frac{\pi}{4}-\left(1-\frac{\pi}{4}\right) \\ &=\frac{\pi}{2}-1. \end{align*}$$ Thus, $$\boxed{\displaystyle \int_{-1}^{1}x\arctan(x)\,dx=\frac{\pi}{2}-1}.$$ The answer is positive, as expected, because the even integrand is nonnegative on $[-1,1]$.