calculate integral of given function

Question

let us consider following integral

while if we calculate from -infinity to plus infinity then it says that

generally it should be 1/infinity +1/infinity right? which should be equal to zero,then why does it says that integral does not converge?

Answer 1

The problem is that there is no primitive of $-\frac1{t^2}$ on all of $\mathbb R$. $\frac1t$ is a primitive of $-\frac1{t^2}$ on $(-\infty,0)$ and on $(0,\infty)$, but not on $\mathbb R$. Thus $$\int_{-\infty}^\infty -\frac1{t^2}\ \mathrm dt \ne \frac1{\infty} - \frac1{-\infty}$$ Looking more closely and using symmetry you can see that actually $$\int_{-\infty}^\infty -\frac1{t^2}\ \mathrm dt = 2\int_0^\infty -\frac1{t^2} \ \mathrm dt = -\infty$$ Where the last integral can be "evaluated" to $-\infty$ by using the primitive on $(0,\infty)$.

A similar thing happens to $\int \frac1t\ \mathrm dt$ with primitive $\ln |t|$ on $(-\infty,0)$ and $(0,\infty)$. Here you can see that $\int_{-1}^1 \frac1t\ \mathrm dt$ is actually undefined although $[\ln |t|]_{-1}^1 = 0$.