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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

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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 $$ \int_0^\infty\frac{\sin x}{1+x}\,dx=1-\int_0^\infty\frac{\cos x}{(1+x)^2}\,dx. $$

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Does that mean the two expressions are unequal in my question ? – Complex analysis Dec 4 '13 at 17:40
They are not equal, and you can check it by yourself! – Mercy King Dec 4 '13 at 17:45
you have forgotten to integrate $\cos x $ again , the first equality after "Notice that " doesn't hold . – Complex analysis Dec 4 '13 at 18:16
@Complexanalysis Are you sure you know how to integrate by parts? – Mercy King Dec 4 '13 at 19:33
Sorry, that was right . That means Baby rudin has a mistake . By the way i still wonder if two integrals are equal does that imply that if one is absolutely convergent then the other is also . I am confused because baby rudin says that even if two integrals are same one may converge absolutely and the other may not . :-/ – Complex analysis Dec 5 '13 at 9:51

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.

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The Babyrudin says that "sometimes" we can apply integration by parts to the indefinite integrals. – Complex analysis Dec 4 '13 at 15:25
You will need to het some feeling for it by tiime. In this case it helps because we will lose $(x+1)^1$ in the denomerator which leads to divergence and get $(1+x)^2$ which will lead to converge. So when you see sometime like trig-function divided by $x$, think about integration by parts, cause this will lead to the square in the denomerator. – user112167 Dec 4 '13 at 15:28
This is not a proof that $\int_0^\infty\frac{\sin x}{1+x}dx does not converge absolutely. – Igor Rivin Dec 4 '13 at 15:46

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...

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+1 Something like $$ \begin{align} \int_0^\infty\left|\frac{\sin(x)}{1+x}\right|\,\mathrm{d}x &\ge\sum_{k=1}^\infty\frac1{1+k\pi}\int_{(k-1)\pi}^{k\pi}|\sin(x)|\,\mathrm{d}x\\ &=\sum_{k=1}^\infty\frac2{1+k\pi}\\ &\ge\frac2\pi\sum_{k=1}^\infty\frac1{k+1} \end{align} $$ – robjohn Dec 4 '13 at 16:06

$\newcommand{\+}{^{\dagger}}% \newcommand{\angles}[1]{\left\langle #1 \right\rangle}% \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}% \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}% \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}% \newcommand{\dd}{{\rm d}}% \newcommand{\down}{\downarrow}% \newcommand{\ds}[1]{\displaystyle{#1}}% \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}% \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}% \newcommand{\fermi}{\,{\rm f}}% \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}% \newcommand{\half}{{1 \over 2}}% \newcommand{\ic}{{\rm i}}% \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow}% \newcommand{\isdiv}{\,\left.\right\vert\,}% \newcommand{\ket}[1]{\left\vert #1\right\rangle}% \newcommand{\ol}[1]{\overline{#1}}% \newcommand{\pars}[1]{\left( #1 \right)}% \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}}% \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}% \newcommand{\sech}{\,{\rm sech}}% \newcommand{\sgn}{\,{\rm sgn}}% \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}}% \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ \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.

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