Integral

**liyi** · December 3rd 2007, 02:21 PM

Evaluate $\int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx$

But without applying $\int\frac1{x^2+1}\,dx=\arctan x+c.$

**colby2152** · December 3rd 2007, 02:42 PM

Let $x = tan(t)$
$x^2 + 1 = tan^2(t) + 1$
$x^2 + 1 = sec^2(t)$

$dx = sec^2(t)dt$
$dt = \frac{dx}{sec^2(t)}$

$\int \frac{1}{x^2 + 1} dx$
$=\int dt$
$=t + c$
$=arctan(x) + c$

**galactus** · December 3rd 2007, 02:48 PM

This is a good case for a little CA since yu can't use arctan.

$\frac{1}{x^{2}+1}=\frac{1}{(x+i)(x-i)}$

The function has a simple pole at i, and the residue is $\frac{1}{2i}$ . Therefore, hence, and whence:

$\int\frac{1}{z^{2}+1}dz=2{\pi}i(\frac{1}{2i})={\pi }$

**topsquark** · December 3rd 2007, 02:55 PM

Originally Posted by liyi

Evaluate $\int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx$

But without applying $\int\frac1{x^2+1}\,dx=\arctan x+c.$

Hmph. I'm not going to do it, but I wonder if you could manage to apply the definition of an integral to this? You know
$\lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{(n \Delta x)^2 + 1} \cdot \Delta x$

The two problems I see are that $\Delta x$ depends on the integration limits, essentially forcing us to calculate $\frac{\infty + \infty}{n}$ , and the resulting series summation that the problem leaves could be a real bear to solve. But these problems might not be insurmountable.

-Dan

**Krizalid** · December 3rd 2007, 03:15 PM

Nice problem!!!

Originally Posted by liyi

Evaluate $\int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx$

Start by showin' that $\int_0^\infty {e^{ - ux} \sin u\,du} = \frac{1} {{x^2 + 1}}.$

We're going to create a double integral. (Of course, the only reason that I'd ever create a double integral in the first place is so I could reverse the integration order.) This yields

$\int_{ - \infty }^{ + \infty } {\frac{1} {{1 + x^2 }}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,du} \,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,dx} \,du} .$

So $\int_{ - \infty }^{ + \infty } {\frac{1} {{1 + x^2 }}\,dx} = 2\int_0^\infty {\frac{{\sin u}} {u}\,du} .$

The right integral can be proven applyin' double integration, but I'll use another method.

Define $f(\varphi ) = \int_0^\infty {\frac{{e^{ - \varphi u} \sin u}} {u}\,du} .$ (Note that we require $f(0).)$

Now $f'(\varphi ) = - \int_0^\infty {e^{ - \varphi u} \sin u\,du} = - \frac{1} {{\varphi ^2 + 1}}.$

Integrate for $\varphi\in(0,\infty),$

$f(\infty ) - f(0) = - \frac{\pi } {2} \implies f(0) = \frac{\pi } {2} + f(\infty ).$

Since $f(\infty)=0,$ we happily get $f(0) = \frac{\pi } {2}.$

--

Yes, I used the fact that $\int\frac1{x^2+1}\,dx=\arctan x+k,$ but the target was to compute the Dirichlet Integral

**ThePerfectHacker** · December 3rd 2007, 05:14 PM

Maybe this will help.

Originally Posted by topsquark

Hmph. I'm not going to do it, but I wonder if you could manage to apply the definition of an integral to this? You know

I cannot image that you can do it. Because integration is defined for bounded sets (or intervals as in this case).

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