Improper Integral using Laplace So I want to calculate the following integral using the laplace transform:
$$\int_0^\infty  \frac{dt}{(t^2+1)^2} $$
I used the theorem:
$$\int_0^\infty f(t)G(t) dt=\int_0^\infty \mathcal{L}[f(t)](x) \space 
\mathcal{L}^{-1} [G(t)](x)dx $$
So if I choose $f(t) = 1 , G(t)=\frac{1}{(t^2+1)^2}$:
$$\mathcal{L}[f(t)]=\frac 1x $$
$$\mathcal{L}^{-1} [G(t)] =\frac 12 (-x\cos{x}+\sin{x})$$
Therefore,
$$\int_0^\infty  \frac{dt}{(t^2+1)^2} = \frac 12 \int_0^\infty \left (-\cos{x}+\frac{\sin{x}}{x} \right ) dx$$
My problem here is with the integral of the cosine, because it does not converge. If I consider it as $0$, I get $\frac{\pi}{4}$ (correct answer), but that is not quite right. I suspect it may have something to do with Cauchy's principal value, but I don't know.
 A: One method of dealing with divergent series defines
$$\sum_{n=0}^\infty ' a_n:=\lim_{n\to\infty} \frac{S_0+S_1+...+S_n}{n+1}$$
where $S_n$ is the $n$th partial sum. This definition assigns the value of any convergent series to itself, which is the Stolz-Cesàro theorem. However, to some oscillatingly divergent series like
$$\sum_{n=0}^\infty (-1)^n$$
it assigns a finite value, such as
$$\sum_{n=0}^\infty' (-1)^n=\frac{1}{2}$$
One might try to do the same thing with integrals by defining
$${\int_0^{\infty}}' f(x)dx:=\lim_{\alpha\to\infty}\frac{1}{\alpha}\int_0^\alpha \frac{1}{x}\int_0^x f(y)\space dydx$$
In which case we have that the integral is assigned to its original value when convergent, and assigned to a "normalized" value when divergent. In your case, we would have
$$\begin{align}
{\int_0^{\infty}}' \cos(x)dx
&=\lim_{\alpha\to\infty}\frac{1}{\alpha}\int_0^\alpha \frac{1}{x}\int_0^x \cos(y)\space dydx\\
&=\lim_{\alpha\to\infty}\frac{1}{\alpha}\int_0^\alpha \frac{\sin(x)}{x}dx\\
&=0
\end{align}$$
...as you suspected. This vanishes since we already know that the Dirichlet Integral converges.
There's just one more thing. You used, in your proof, the fact that
$$\int_0^\infty f(x)g(x)dx=\int_0^\infty (\mathcal L f)(x)(\mathcal L^{-1} g)(x)dx$$
all you have to do now is that this theorem still holds for our normalized definition of the integral. Shouldn't be too hard, provided you know the proof of the original theorem.
Now, might I suggest a few other methods for evaluating this troublesome integral?
DIFFERENTIATION UNDER THE INTEGRAL: We already know that
$$\int_0^\infty \frac{dt}{t^2+a^2}=\frac{\pi}{2}\frac{1}{a}$$
and so by differentiating both sides with respect to $a$ and moving the derivative into the integral, and then dividing both sides by $-2a$, we have that
$$\int_0^\infty \frac{dt}{(t^2+a^2)^2}=\frac{\pi}{4}\frac{1}{a^3}$$
and by letting $a=1$:
$$\int_0^\infty \frac{dt}{(t^2+1)^2}=\frac{\pi}{4}$$
BETA FUNCTION: Make the substitution $t\to \sqrt t$ to get
$$\frac{1}{2}\int_0^\infty \frac{dt}{\sqrt{t}(t+1)^2}$$
then make the substitution $t\to \frac{t}{1-t}$ to get
$$\frac{1}{2}\int_0^1 \frac{\sqrt{1-t}\space dt}{\sqrt{t}}$$
Which, using the Beta function, is equal to
$$\frac{1}{2}\frac{\Gamma(1/2)\Gamma(3/2)}{\Gamma(2)}=\frac{\pi}{4}$$
as desired.
RESIDUE THEOREM: Let your contour $\gamma$ be a semicircular contour with radius $r$ centered at the origin reaching into the first and fourth quadrants. By parametrizing, the integral of your function over this contour is
$$\oint_\gamma \frac{dt}{(t^2+1)^2}=\int_{-r}^r \frac{dt}{(t^2+1)^2}+\int_{0}^{\pi} \frac{ire^{i\theta}\space d\theta}{(r^2e^{2i\theta}+1)^2}$$
Letting $r$ go to infinity causes the second integral to vanish, leaving us with
$$\oint_\gamma \frac{dt}{(t^2+1)^2}\approx\int_{-r}^r \frac{dt}{(t^2+1)^2}$$
The only pole of the integrand within the contour is at $t=i$ with an order of $2$, which has a residue of
$$\lim_{t\to i}\frac{d}{dt}\frac{(t-i)^2}{(t^2+1)^2}=-\frac{i}{4}$$
Thus, by the Residue Theorem, letting $r\to\infty$, we have
$$\int_{-\infty}^\infty \frac{dt}{(t^2+1)^2}=\frac{\pi}{2}$$
or
$$\int_{0}^\infty \frac{dt}{(t^2+1)^2}=\frac{\pi}{4}$$
ANTIDERIVATIVE: I've saved the best for last. The integrand is, after all, antidifferentiable:
$$\int \frac{dt}{(t^2+1)^2}=\frac{1}{2}\frac{t}{t^2+1}+\frac{\arctan(t)}{2}$$
So we have
$$\int_0^\infty \frac{dt}{(t^2+1)^2}=\lim_{\alpha\to\infty} \bigg[\frac{1}{2}\frac{t}{t^2+1}+\frac{\arctan(t)}{2}\bigg]_0^\alpha=\frac{\pi}{4}$$
A: The solution can be much shorter without using Laplas transform:
Let $t=\tan x$; we have:
$1+t^2=1+\tan^2 x=\frac{1}{\cos^2 x}$and $dt =(1+\tan^2 x)dx$
Sustituting we get:
$\int^∞_0 \frac{dt}{(1+t^2)^2}=\int^{\pi/2}_0 \cos^2 x=[\frac {x}{2}+\frac{\sin 2x}{4}]^{\pi/2}_0=\frac{\pi}{4} $
