As the title says, does this improper integral $\int_0^{\infty}\frac{\sin{x}}{x(1+x)}dx$ exist?

And if so, how can I prove it?

It looks a lot like the sine integral $Si(x)=\int_0^{x}\frac{\sin{t}}{t}dt$, but I don't know how further show the correlation.

Any advice would be very helpful!


It exists by Dirichlet's test (integral version), since $\sin(x)$ has a bounded primitive and $\frac{1}{x(1+x)}$ is decreasing towards zero. There are no integrability issues in a right neighbourhood of the origin, since $\frac{\sin x}{x}$ is an entire function. Moreover, by the Laplace transform:

$$ \int_{0}^{+\infty}\frac{\sin x}{x(1+x)}\,dx = \int_{0}^{+\infty}\frac{1-e^{-s}}{1+s^2}\,ds=\frac{\pi}{2}-\int_{0}^{+\infty}\frac{dt}{(1+t^2)\,e^t}. $$

This transformation removes the oscillations from the integrand function and allows us to provide tight approximations for the value of the integral, that can be expressed in terms of the sine and cosine integral. For instance, the Cauchy-Schwarz inequality gives:

$$ \int_{0}^{+\infty}\frac{dt}{(1+t^2)e^t}\leq \sqrt{\left(\int_{0}^{+\infty}\frac{dt}{(4+t)(1+t^2)^2}\right)\cdot\left(\int_{0}^{+\infty}(t+4)e^{-2t}\,dt\right)}$$ and so we get: $$ \int_{0}^{+\infty}\frac{dt}{(1+t^2)e^t}\leq\sqrt{\frac{9 (38 \pi-17 -2 \log 4)}{2312}}=\color{red}{0.62}269345635\ldots $$ where the actual value of the last integral is $\color{red}{0.62}14496242358\ldots $
A tight lower bound can be achieved in the same way: $$ \int_{0}^{+\infty}\frac{dt}{(1+t^2)e^t} = 1-\int_{0}^{+\infty}\frac{t^2\,dt}{(1+t^2)e^t} $$ and: $$ \int_{0}^{+\infty}\frac{t^2 dt}{(1+t^2)e^t}\leq \sqrt{\left(\int_{0}^{+\infty}\frac{t^2\,dt}{(1+t)(1+t^2)^2}\right)\cdot\left(\int_{0}^{+\infty}t^2(t+1)e^{-2t}\,dt\right)}$$ gives: $$ \int_{0}^{+\infty}\frac{dt}{(1+t^2)e^t}\geq 1-\sqrt{\frac{5}{32}}>\color{red}{0.6}.$$


Split it into two pieces:

$$\int \limits _0 ^1 \frac {\sin x} {x(1+x)} \ \Bbb d x + \int \limits _1 ^\infty \frac {\sin x} {x(1+x)} \ \Bbb d x .$$

For the first one, notice that the only problem is in $0$, but since $\lim \limits _{x \to 0} \frac {\sin x} x = 1$, the integrand essentially behaves like $\frac 1 {1+x}$, which is integrable on $[0,1]$.

For the second one, notice that

$$0 \le \left| \int \limits _1 ^\infty \frac {\sin x} {x(1+x)} \ \Bbb d x \right| \le \int \limits _1 ^\infty \left| \frac {\sin x} {x(1+x)} \right| \ \Bbb d x \le \int \limits _1 ^\infty \frac 1 {x(1+x)} \ \Bbb d x \le \int \limits _1 ^\infty \frac 1 {x^2} \ \Bbb d x$$

which is finite, so the second term is integrable, too.

Hence, the original integral is convergent.

(The first integral is an improper integral of the second type, while the second integral is an improper integral of the first type. For the first one I have used the limit comparison test, while for the second one I have used the inequality comparison test - they are the same ones as the ones used for series, but with integrals instead of sums.)

Alternatively, if you already feel comfortable with the integral $\int \limits _0 ^\infty \frac {\sin x} x \ \Bbb d x$, as your post suggests, then notice that

$$\int \limits _0 ^\infty \frac {\sin x} {x(1+x)} \ \Bbb d x = \int \limits _0 ^\infty \frac {\sin x} x \ \Bbb d x - \int \limits _0 ^\infty \frac {\sin x} {1+x} \ \Bbb d x$$

and the second integral is convergent with the same type of argument as for the first one - for instance, using Abel-Dirichlet's theorem (notice that the second integral, unlike the first one, has problems only at $\infty$).


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