Existence of $\int_{-a}^{a}\frac{1}{t}dt$ I've never understood why $\int_{-a}^{a}\frac{1}{t}dt$ isn't $0$. The function $\frac1t$ is odd so the area under the curve on the $t=(0,a]$ interval should cancel out with the area on the $t=[-a,0)$ interval. It shouldn't matter whether it's infinite area though right?
 A: If you have to assign a meaningful value to it, then use the Cauchy Principal Value, which is obtained by direct integration.
Similar to how physics uses $1+2+3...=-\frac{1}{12}$. 
But if you want something intuitive, like the rate of water pouring into a glass, pouring infinite amount of water, and taking an infinite amount of water out right after... yeah, that doesn't quite make sense, because the integral doesn't converge when broken apart. 
A: This is called the Cauchy principal value, and it has its uses, as long as you understand its drawbacks.
The big thing you lose is the change of variables.
Let $f(x)=\frac{1}{x}$ and $$g(x)=\begin{cases}\frac{1}{x}&x<0\\\frac{n}{x}&x>0\end{cases}$$
If you want $\int_{-1}^{1} g(x)\,dx$, you might let $u=x^n$ and $du=nx^{n-1}$
You might try the substitution $$u=\begin{cases}x^n&x>0\\x&x<0\end{cases}$$.
So $\frac{du}{u}=\frac{n\,dx}{x}$ for all $x$, but you the Cauchy principal value for our integral is not the same (the value is infinite for the CPV of $\int_{-1}^1 g(x)\,dx$.)

A similar risk can be seen if you say it is "obvious" that $$\int_{-\infty}^{\infty} x\,dx=0,$$ but also, it is "obvious" that:
$$\int_{-\infty}^{\infty} (x+1)\,dx = \int_{-\infty}^{\infty} x\,dx$$
A: The integral $\int_{-a}^{a}\frac{1}{t}dt$ is convergent $ \iff$ the integrals $\int_{-a}^{0}\frac{1}{t}dt$ and $\int_{0}^{a}\frac{1}{t}dt$ are both convergent.
But the integrals $\int_{-a}^{0}\frac{1}{t}dt$ and $\int_{0}^{a}\frac{1}{t}dt$ are both divergent !
A: Note that the integral does not make sense at all when viewed as a Riemann integral because the integrand $f(x) =1/x$ is unbounded on the interval of integration $[-a, a] $. There is way to deal with unbounded functions via improper Riemann integrals and it works by assuming that there are isolated points in whose neighborhood the function is unbounded.
In the current case such a point is $0$ and then the idea is to split the interval of integration into $[-a, 0]$ and $[0,a]$ and deal with both integrals $$\int_{-a} ^{0}\frac{dx}{x},\int_{0}^{a}\frac{dx}{x}$$ as improper Riemann integrals. Unfortunately both these integrals diverge and hence this approach also fails so that original integral does not make sense even as improper Riemann integral. However there is still some hope and the integral has Cauchy principal value $0$ which sort of meets your expectation about integral of odd functions over intervals of type $[-a, a] $.
