Evaluate $\displaystyle \int_{-\infty}^{+\infty}\frac1{x^2+1}\,dx$
But without applying $\displaystyle \int\frac1{x^2+1}\,dx=\arctan x+c.$
Let $\displaystyle x = tan(t)$
$\displaystyle x^2 + 1 = tan^2(t) + 1$
$\displaystyle x^2 + 1 = sec^2(t)$
$\displaystyle dx = sec^2(t)dt$
$\displaystyle dt = \frac{dx}{sec^2(t)}$
$\displaystyle \int \frac{1}{x^2 + 1} dx$
$\displaystyle =\int dt$
$\displaystyle =t + c$
$\displaystyle =arctan(x) + c$
This is a good case for a little CA since yu can't use arctan.
$\displaystyle \frac{1}{x^{2}+1}=\frac{1}{(x+i)(x-i)}$
The function has a simple pole at i, and the residue is $\displaystyle \frac{1}{2i}$. Therefore, hence, and whence:
$\displaystyle \int\frac{1}{z^{2}+1}dz=2{\pi}i(\frac{1}{2i})={\pi }$
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
$\displaystyle \lim_{n \to \infty} \sum_{i = 1}^n \frac{1}{(n \Delta x)^2 + 1} \cdot \Delta x$
The two problems I see are that $\displaystyle \Delta x$ depends on the integration limits, essentially forcing us to calculate $\displaystyle \frac{\infty + \infty}{n}$, and the resulting series summation that the problem leaves could be a real bear to solve. But these problems might not be insurmountable.
-Dan
Nice problem!!!
Start by showin' that $\displaystyle \int_0^\infty {e^{ - ux} \sin u\,du} = \frac{1}
{{x^2 + 1}}.$
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
$\displaystyle \int_{ - \infty }^{ + \infty } {\frac{1}
{{1 + x^2 }}\,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,du} \,dx} = 2\int_0^\infty {\int_0^\infty {e^{ - ux} \sin u\,dx} \,du} .$
So $\displaystyle \int_{ - \infty }^{ + \infty } {\frac{1}
{{1 + x^2 }}\,dx} = 2\int_0^\infty {\frac{{\sin u}}
{u}\,du} .$
The right integral can be proven applyin' double integration, but I'll use another method.
Define $\displaystyle f(\varphi ) = \int_0^\infty {\frac{{e^{ - \varphi u} \sin u}}
{u}\,du} .$ (Note that we require $\displaystyle f(0).)$
Now $\displaystyle f'(\varphi ) = - \int_0^\infty {e^{ - \varphi u} \sin u\,du} = - \frac{1}
{{\varphi ^2 + 1}}.$
Integrate for $\displaystyle \varphi\in(0,\infty),$
$\displaystyle f(\infty ) - f(0) = - \frac{\pi }
{2} \implies f(0) = \frac{\pi }
{2} + f(\infty ).$
Since $\displaystyle f(\infty)=0,$ we happily get $\displaystyle f(0) = \frac{\pi }
{2}.$
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Yes, I used the fact that $\displaystyle \int\frac1{x^2+1}\,dx=\arctan x+k,$ but the target was to compute the Dirichlet Integral