# 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

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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 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 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

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