1. ## Integrals

Evaluate

$\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$

$\displaystyle \int_0^1\frac1{(x^2-x+1)(e^{2x-1}+1)}\,dx$

2. Originally Posted by liyi
Evaluate
$\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
$\displaystyle 1 - \cos x = 2\sin^2 \frac{x}{2}$
Thus,
$\displaystyle 2\int_0^{\infty} \frac{\sin^2 \frac{x}{2}}{x^2} dx$
Let $\displaystyle t=x/2$,
$\displaystyle 2\int_0^{\infty} \frac{\sin^2 t}{4t^2} \cdot 2dt = \frac{\pi}{2}$

3. Originally Posted by liyi
Evaluate

$\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
Solution 2. In my other post I used that fact that $\displaystyle \int_0^{\infty} \frac{\sin^2 t}{t^2}~dt = \frac{\pi}{2}$ (it was proven on this forum before).

Here is a method just using the fact that $\displaystyle \int_0^{\infty} \frac{\sin x}{x} dx = \frac{\pi}{2}$.

Note that,
$\displaystyle \int_0^{\infty} \frac{1-\cos x}{x^2} dx = \int_0^{\infty} \int_0^y \frac{\sin y}{x^2} dy~dx =$$\displaystyle \int_0^{\infty} \int_y^{\infty} \frac{\sin y}{x^2} dx~dy = \int_0^{\infty} \frac{\sin y}{y} dy = \frac{\pi}{2}$

4. Originally Posted by liyi
Evaluate

$\displaystyle \int_0^\infty\frac{1-\cos x}{x^2}\,dx$
Here's another proof.

We'll use the following fact $\displaystyle \int_0^\infty\frac{\sin x}x\,dx=\frac\pi2.$ (Proven here.)

Now $\displaystyle \frac{{1 - \cos x}} {x} = \int_0^1 {\sin (ux)\,du} .$

The integral becomes to $\displaystyle \int_0^\infty {\int_0^1 {\frac{{\sin (ux)}} {x}\,du} \,dx} = \int_0^1 {\int_0^\infty {\frac{{\sin (ux)}} {x}\,dx} \,du} .$

For the inner integral, substitute $\displaystyle \varphi=ux,$

$\displaystyle \int_0^\infty {\frac{{\sin (ux)}} {x}\,dx} = \frac{1} {u}\int_0^\infty {\frac{{u\sin \varphi }} {\varphi }\,d\varphi } = \frac{\pi } {2}.$

Finally

$\displaystyle \int_0^\infty {\frac{{1 - \cos x}} {{x^2 }}\,dx} = \frac{\pi } {2}\int_0^1 {du} = \frac{\pi } {2}\,\blacksquare$

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A related problem.

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As for $\displaystyle \int_0^\infty {\frac{{\sin ^2 x}} {{x^2 }}\,dx}$ can be found here.

5. Nice problem

Originally Posted by liyi
$\displaystyle \int_0^1\frac1{(x^2-x+1)(e^{2x-1}+1)}\,dx$
Define $\displaystyle u=2x-1,$

$\displaystyle \int_0^1 {\frac{1} {{\left( {x^2 - x + 1} \right)\left( {e^{2x - 1} + 1} \right)}}\,dx} = \int_{ - 1}^1 {\frac{2} {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} .$

Substitute $\displaystyle u=-\varphi,$

$\displaystyle \int_{ - 1}^1 {\frac{2} {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2} {{\left( {\varphi ^2 + 3} \right)\left( {e^{ - \varphi } + 1} \right)}}\,d\varphi ,}$ so

$\displaystyle \tau = \int_{ - 1}^1 {\frac{2} {{\left( {u^2 + 3} \right)\left( {e^u + 1} \right)}}\,du} = \int_{ - 1}^1 {\frac{2} {{\left( {u^2 + 3} \right)\left( {e^{ - u} + 1} \right)}}\,du} .$

And finally

$\displaystyle 2\tau = \int_{ - 1}^1 {\frac{2} {{u^2 + 3}}\,du} = \frac{{2\pi }} {{3\sqrt 3 }}\,\therefore \,\tau = \frac{\pi } {{3\sqrt 3 }}.$