Integral using residues

Question

Can't solve $\int_{0}^{+\infty}\frac{x-\sin{x}}{x^3}dx$. Try to use $\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$ and $\sin{x}=\Im{e^{ix}}$ but also can't solve this integral. Any suggestions how to solve it? Thanks in advance.

Edit:

If i use: $\sin{x}=\Im{e^{ix}}$: $\int_{0}^{+\infty}\frac{x-\sin{x}}{x^3}dx$ = $\int_{0}^{+\infty}\frac{x-\Im{e^{ix}}}{x^3}dx$ = $\Im{\int_{0}^{+\infty}\frac{x-e^{ix}}{x^3}dx}=\Im{I}$

Then calculate $I$ using residue theorem: $I=2\pi i (res_{z=0}{f(z)})=2\pi i \frac{1}{2}=\pi i \Rightarrow$

$\Im{I}=0$ and that isn't correct answer, because correct answer is $\pi/4$.

Answer 1

Posting an entirely different approach based on the above statement

Any suggestions how to solve it?

Using integration by parts,

$$\int\frac{x-\sin x}{x^3}dx = (x-\sin x)\frac{-1}{2x^2} +\frac{1}{2}\int\frac{1-\cos x}{x^2}dx$$ $$ = -\frac{1}{2}\frac{x-\sin x}{x^2} + \frac{1}{2}(1-\cos x)\frac{-1}{x} + \frac{1}{2}\int\frac{\sin x}{x}dx$$ $$ = \frac{-x+\sin x -x + x\cos x}{2x^2} + \frac{1}{2}\int\frac{\sin x}{x}dx$$

Hence $$\int_0^\infty\frac{x-\sin x}{x^3}dx = \left.\left(-\frac{1}{x} + \frac{\cos x}{2x} + \frac{\sin x}{2x^2}\right)\right|_0^\infty + \frac{\pi}{4}$$

Can you evaluate the limit and work out the rest?

Answer 2

Another way to do it :

Define on $ \mathbb{R}_{+} $, a function $ f $ as follows : $$ \left(\forall t\in\mathbb{R}_{+}\right),\ f\left(t\right)=\int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{3}}\,\mathrm{e}^{-xt}\,\mathrm{d}x} $$

Since $ \left(\forall x\in\mathbb{R}_{+}^{*}\right),\ t\mapsto\frac{x-\sin{x}}{x^{3}}\,\mathrm{e}^{-xt} $ is continuous on $ \mathbb{R}_{+} $, and $ \left(\forall a,b\in\mathbb{R}_{+}\right)\left(\forall\left(t,x\right)\in\left[a,b\right]\times\mathbb{R}_{+}^{*}\right),\ \frac{x-\sin{x}}{x^{3}}\,\mathrm{e}^{-xt}\leq\frac{x-\sin{x}}{x^{3}} $, with $ \varphi:x\mapsto\frac{x-\sin{x}}{x^{3}} $ being integrable on $ \mathbb{R}_{+}^{*} $, we conclude that $ f $ is continuous on $ \mathbb{R}_{+} \cdot $

Now since : $$ \left(\forall x\in\mathbb{R}_{+}^{*}\right),\ \frac{x-\sin{x}}{x^{3}}=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{d}y} $$

We have, for any $ t\in\mathbb{R}_{+}^{*} $, the following : \begin{aligned}f\left(t\right)&=\frac{1}{2}\int_{0}^{+\infty}{\int_{0}^{1}{\left(1-y\right)^{2}\cos{\left(xy\right)}\,\mathrm{e}^{-xt}\,\mathrm{d}y}\,\mathrm{d}x}\\ &=\frac{1}{2}\int_{0}^{1}{\left(1-y\right)^{2}\int_{0}^{+\infty}{\cos{\left(xy\right)}\,\mathrm{e}^{-xt}\,\mathrm{d}x}\,\mathrm{d}y}\\ &=\frac{t}{2}\int_{0}^{1}{\frac{\left(1-y\right)^{2}}{t^{2}+y^{2}}\,\mathrm{d}y}\\ &=\frac{t}{2}+\frac{1-t^{2}}{2}\int_{0}^{1}{\frac{\frac{1}{t}}{1+\left(\frac{y}{t}\right)^{2}}\,\mathrm{d}t}-\frac{t}{2}\int_{0}^{1}{\frac{2y}{t^{2}+y^{2}}\,\mathrm{d}y}\\ f\left(t\right)&=\frac{t}{2}+\frac{1-t^{2}}{2}\arctan{\left(\frac{1}{t}\right)}-\frac{t}{2}\ln{\left(1+t\right)}+t\ln{t}\end{aligned}

Since $ \lim\limits_{t\to 0}{f\left(t\right)}=\frac{\pi}{4} $, and $ f $ is continuous on $ \mathbb{R}_{+} $, we have $ f\left(0\right)=\frac{\pi}{4} $, and thus : $$ \int_{0}^{+\infty}{\frac{x-\sin{x}}{x^{3}}\,\mathrm{d}x}=\frac{\pi}{4} $$