Analysis: Prove that this improper integral diverges but the limit as $t \rightarrow \infty=\pi$?

Question

Question:

Show that the improper integral $$\int_{-\infty}^{\infty}\frac{1+x}{1+x^2}dx$$ diverges but that $$\lim_{t\rightarrow\infty}\int_{-t}^{t}\frac{1+x}{1+x^2}dx=\pi.$$

Answer 1

It is easy to check that $\frac{1 + x}{1 + x^2}$ is a bounded function on $\left (-\infty, \infty \right)$, in particular it is bounded on every bounded, closed subinterval. This is an improper integral because the interval of integration $\left (-\infty, \infty \right)$ is infinite. We first need to write$\int_{-\infty}^{\infty}\frac{1+x}{1+x^2}dx$ as the sum of two improper integrals, for instance as
 $$\int_{-\infty}^{\infty}\frac{1+x}{1+x^2}dx= \int_{-\infty}^{0}\frac{1+x}{1+x^2}dx+\int_{0}^{\infty}\frac{1+x}{1+x^2}dx,$$

and then evaluate the two resulting improper integral separately. So,

$$\int_{0}^{\infty}\frac{1+x}{1+x^2}dx=\lim_{M\rightarrow\infty}\int_{0}^{M}\frac{1+x}{1+x^2}dx$$

$$=\lim_{M\rightarrow\infty} \left [ \int_{0}^{M}\frac{1}{1+x^2}dx+\int_{0}^{M}\frac{x}{1+x^2}dx \right ]$$

$$=\lim_{M\rightarrow\infty} \left [\left (\arctan(M) - \arctan(0)\right)+ \left ( \frac{1}{2}\ln \left (1+M^2 \right) \right) \right] = \bf{\infty},$$

since $\lim_{M\rightarrow\infty}\ln(1+M^2) = \infty$. Therefore the improper integral $\int_{0}^{\infty}\frac{1+x}{1+x^2}dx$ diverges, and hence the original improper integral $\int_{-\infty}^{\infty}\frac{1+x}{1+x^2}dx$ diverges.

However, when we evaluate $\lim_{t\rightarrow \infty}\int_{-t}^{t}\frac{1+x}{1+x^2}dx$ we get

$$\lim_{t \rightarrow \infty}\int_{-t}^{t} \frac{1+x}{1+x^2}dx=\lim_{t \rightarrow \infty} \left [\int_{-t}^{t} \frac{1}{1+x^2}dx+\int_{-t}^{t} \frac{x}{1+x^2}dx \right]$$

$$=\lim_{t \rightarrow \infty} \left [\left (\arctan(t) - \arctan(-t) \right ) + \frac{1}{2} \left ( \ln(1+t^2) -\ln(1+(-t)^2) \right ) \right ]$$

$$=\lim_{t \rightarrow \infty} 2 \space \arctan(t) = 2 \frac{\pi}{2} = \pi$$

and so $lim_{t \rightarrow \infty} \int_{-t}^{t} \frac{1+x}{1+x^2}dx$ converges. We have used the fact that inverse tangent is an odd function, that is $\arctan(-t) = -\arctan(t)$.

Answer 2

Hint: $$\frac{1+x}{1+x^2} \operatorname*{\sim}_{x\to\infty} \frac{1}{x}$$ which is not integrable around $\infty$. But $$ \int_{-t}^{t} \frac{1+x}{1+x^2}dx = \int_{0}^{t} \frac{1+x}{1+x^2}dx+\int_{0}^{t} \frac{1-x}{1+x^2}dx = 2\cdot \int_{0}^{t} \frac{1}{1+x^2}dx. $$

Answer 3

$$\int\frac{1+x}{1+x^2}\space\text{d}x=\int\left(\frac{x}{x^2+1}+\frac{1}{x^2+1}\right)\space\text{d}x=$$ $$\int\frac{x}{x^2+1}\space\text{d}x+\int\frac{1}{x^2+1}\space\text{d}x=$$

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Substitute $u=x^2+1$ and $\text{d}u=2x\space\text{d}x$:

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$$\frac{1}{2}\int\frac{1}{u}\space\text{d}x+\int\frac{1}{x^2+1}\space\text{d}x= \frac{\ln\left|u\right|}{2}+\int\frac{1}{x^2+1}\space\text{d}x=$$ $$\frac{\ln\left|u\right|}{2}+\arctan\left(x\right)+\text{C}=\frac{\ln\left|x^2+1\right|}{2}+\arctan\left(x\right)+\text{C}$$

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Now set the boundaries:

$$\lim_{t\to\infty}\int_{-t}^{t}\frac{1+x}{1+x^2}\space\text{d}x=\lim_{t\to\infty}\left[\frac{\ln\left|x^2+1\right|}{2}+\arctan\left(x\right)\right]_{-t}^{t}=$$ $$\lim_{t\to\infty}\left(\left(\frac{\ln\left|t^2+1\right|}{2}+\arctan\left(t\right)\right)-\left(\frac{\ln\left|(-t)^2+1\right|}{2}+\arctan\left(-t\right)\right)\right)=$$ $$\lim_{t\to\infty}\left(2\arctan(t)\right)=2\lim_{t\to\infty}\arctan(t)=2\cdot\frac{\pi}{2}=\frac{2}{2}\cdot\pi=\pi$$