Integrals

**liyi** · December 6th 2007, 06:02 PM

Evaluate

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

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

**ThePerfectHacker** · December 6th 2007, 06:45 PM

Originally Posted by liyi

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

$1 - \cos x = 2\sin^2 \frac{x}{2}$
Thus,
$2\int_0^{\infty} \frac{\sin^2 \frac{x}{2}}{x^2} dx$
Let $t=x/2$ ,
$2\int_0^{\infty} \frac{\sin^2 t}{4t^2} \cdot 2dt = \frac{\pi}{2}$

**ThePerfectHacker** · December 6th 2007, 06:53 PM

Originally Posted by liyi

Evaluate

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

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

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

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

**Krizalid** · December 7th 2007, 05:04 AM

Originally Posted by liyi

Evaluate

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

Here's another proof.

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

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

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

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

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

Finally

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

--

A related problem.

--

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

**Krizalid** · December 7th 2007, 05:29 AM

Nice problem

Originally Posted by liyi

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

Define $u=2x-1,$

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

Substitute $u=-\varphi,$

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

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

And finally

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

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