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\noindent {\bf Question}

\noindent Show that $\int_{-\infty}^\infty (1+x){\rm d}x/(1+x^2)$
diverges, but that $\lim_{t\rightarrow\infty} \int_{-t}^t
(1+x){\rm d}x/(1+x^2) =\pi$.

\medskip

\noindent {\bf Answer}

\noindent We first need to write $\int_{-\infty}^\infty (1+x){\rm
d}x/(1+x^2)$ as the sum of two improper integrals, for instance
\[ \int_{-\infty}^\infty \frac{1+x}{1+x^2}\: {\rm d}x =
\int_{-\infty}^0 \frac{1+x}{1+x^2}\: {\rm d}x + \int_0^\infty
\frac{1+x}{1+x^2}\: {\rm d}x, \] and then evaluate the two
resulting improper integrals separatedly. So,
\begin{eqnarray*}
\int_0^\infty \frac{1+x}{1+x^2}\: {\rm d}x &amp; = &amp;
\lim_{M\rightarrow\infty} \int_0^M \frac{1+x}{1+x^2}\: {\rm d}x \\
 &amp; = &amp; \lim_{M\rightarrow\infty} \left[ \int_0^M \frac{1}{1+x^2}\:
{\rm d}x + \int_0^M \frac{x}{1+x^2}\: {\rm d}x \right] \\
 &amp; = &amp; \lim_{M\rightarrow\infty} \left[ (\arctan(M) - \arctan(0))
+ \left( \frac{1}{2} \ln(1+M^2) - \frac{1}{2}\right) \right] =
\infty,
\end{eqnarray*}
since $\lim_{M\rightarrow\infty} \ln(1+M^2) =\infty$, and so the
original improper integral $\int_{-\infty}^\infty (1+x){\rm
d}x/(1+x^2)$ diverges.

\medskip
\noindent However, when we evaluate $\lim_{t\rightarrow\infty}
\int_{-t}^t (1+x){\rm d}x/(1+x^2)$, we get
\begin{eqnarray*}
\lim_{t\rightarrow\infty} \int_{-t}^t \frac{1+x}{1+x^2}\: {\rm d}x
&amp; = &amp; \lim_{t\rightarrow\infty} \left[ \int_{-t}^t
\frac{1}{1+x^2}\: {\rm d}x + \int_{-t}^t \frac{x}{1+x^2}\: {\rm
d}x \right] \\ &amp; = &amp; \lim_{t\rightarrow\infty} \left[ (\arctan(t)
-\arctan(-t)) + \frac{1}{2}\left( \ln(1 +t^2) - \ln(1 + (-t)^2)
\right) \right] \\ &amp; = &amp; \lim_{t\rightarrow\infty} 2\arctan(t)
=2\frac{\pi}{2} =\pi,
\end{eqnarray*}
and so $\lim_{t\rightarrow\infty} \int_{-t}^t (1+x){\rm
d}x/(1+x^2)$ converges.  (Here, we use that $\arctan(-t)
=-\arctan(t)$.)


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