Equal integral but only one of them converges absolutely . Consider the following integral 
$$\int_0 ^\infty \frac{\sin x}{1+x} \, dx.$$
By integration by parts we get $$\int_0^\infty \frac{\cos x}{(1+x)^2}\,dx.$$
But according to Rudin , one of them is absolutely convergent and the other isn't. How do i prove it.
$$\int_0^\infty  \left| \frac{\cos x}{ (1+x)^2}\right|\,dx \le  \int_0^\infty \frac{1}{(1+x)^2} \,dx < \infty $$ This one is quite clear . 
Another question is what is the necessary condition on a integral so that we can do INTEGRATION BY PARTS. 
I find it amusing by the fact that even though both the integrals are same but one of them converges absolutely and the other doesn't . 
Thanks 
 A: We can always do integration by parts. The reason that the second integral is absolute convergent is because of the square in the denomerator. $\int_0^{\infty} \frac{1}{1+x} dx$ does not converge, while $\int_0^{\infty} \frac{1}{(1+x)^2} dx$ does. You can show that the first integral does not converge bu making the substitution $u=x+1$. This would give a natural logarithm function. The rest I leave up to you.
A: To show that the first integral does not converge absolutely, take absolute values, and then note that $\int_0^\infty \frac{|\sin x|}{1+x} dx \geq \sum_{k=1}^\infty \int_{\pi/2 + 2\pi k - \epsilon}^{\pi/2 + 2\pi k + \epsilon}\frac{|\sin x|}{1+x} dx.$ The $k$-th integral in the sum is bounded below by $c/k,$ for some positive constant $c,$ since $\sin \pi/2 = 1,$ so the sum diverges by comparison with the harmonic series...
A: The integral $\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx$ diverges. In fact, we have
\begin{eqnarray}
\int_0^\infty\left|\frac{\sin x}{1+x}\right|\,dx&=&\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left|\frac{\sin x}{1+x}\right|\,dx=\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin(x+k\pi)}{1+k\pi+x}\right|\,dx\\
&=&\sum_{k=0}^\infty\int_0^\pi\left|\frac{\sin x}{1+k\pi+x}\right|\,dx
=\sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+k\pi+x}\,dx\\
&\ge& \sum_{k=0}^\infty\int_0^\pi\frac{\sin x}{1+(k+1)\pi}\,dx=\sum_{k=0}^\infty\frac{2}{1+(k+1)\pi}=\infty.
\end{eqnarray}
Hence the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ isn't absolutely convergent.In contrast we have
$$
\int_0^\infty\left|\frac{\cos x}{(1+x)^2}\right|\,dx\le\int_0^\infty\frac{1}{(1+x)^2}\,dx=1,
$$
i.e. the integral $\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx$ is absolutely convergent.
However, the integral $\int_0^\infty\frac{\sin x}{1+x}\,dx$ does converge, because we have
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=\sum_{k=0}^\infty(-1)^k\int_0^\pi\frac{\sin x}{1+x+k\pi}\,dx=:\sum_{k=0}^\infty(-1)^ka_k,
$$
where the sequence $(a_k)$ satisfies the following:
\begin{eqnarray}
a_{k+1}-a_k&=&\int_0^\pi\sin x\left(\frac{1}{1+x+k\pi+\pi}-\frac{1}{1+x+k\pi}\right)\,dx\\
&=&-\int_0^\pi\frac{\pi\sin x}{(1+x+k\pi)(1+x+k\pi+\pi)}\,dx\le 0,
\end{eqnarray}
and
$$
0\le a_k\le \int_0^\pi\frac{\sin x}{1+k\pi}\,dx=\frac{2}{1+k\pi}\to 0.
$$
Notice that for every $\theta>0$ we have
$$
\int_0^\theta\frac{\sin x}{1+x}\,dx=-\frac{\cos x}{1+x}\Big|_0^\theta-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx=1-\frac{\cos\theta}{1+\theta}-\int_0^\theta\frac{\cos x}{(1+x)^2}\,dx.
$$
It follows that
$$
\int_0^\infty\frac{\sin x}{1+x}\,dx=1-\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx.
$$
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
\color{#00f}{\large\int_{0}^{\infty}{\sin\pars{x} \over 1 + x}\,\dd x}&=
\int_{1}^{\infty}{\sin\pars{x - 1} \over x}\,\dd x
=
\int_{1}^{\infty}{\sin\pars{x}\cos\pars{1} - \cos\pars{x}\sin\pars{1} \over x}\,\dd x
\\[3mm]&=\cos\pars{1}\bracks{%
\int_{0}^{\infty}{\sin{x} \over x}\,\dd x - \int_{0}^{1}{\sin{x} \over x}\,\dd x}
+\sin\pars{1}\bracks{-\int_{1}^{\infty}{\cos\pars{x} \over x}\,\dd x}
\\[3mm]&=
\color{#00f}{\large{\pi \over 2}\,\cos\pars{1} - \cos\pars{1}{\rm Si}\pars{1} + \sin\pars{1}{\rm Ci}\pars{1}} \approx 0.6215
\end{align}
where ${\rm Si}$ and ${\rm Ci}$ are the Sine Integral Function and the
Cosine Integral Function, respectively.
