Integral $\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx$ $$\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx$$
I subbed in $x=a \tan\theta$ and ended up with $\ln|\sec\theta+\tan\theta|$. Is this correct?
Thanks.
 A: Here,  you can just use the fact that the integrand is very close to $|\frac 1 x |$.  Specifically,  break up the integral into the regions from $-\infty$ to $|a|$, then from $|a|$ to $\infty$.   Both are nonnegative, so if the latter is unbounded above, so is the whole function.   Now,  in the second region,  $x^2\ge a^2$,  so $\sqrt {x^2 +x^2}>\sqrt {x^2 +a^2}$,  hence $\frac 1 {\sqrt {2x^2}}<\frac 1 {\sqrt {x^2 +a^2}}$,  thus your integral is bigger than $\int _{|a|} ^\infty \frac 1 {\sqrt 2}\cdot \frac 1 xdx=\infty $
So by comparison, your integral diverges
A: Yes, now resub $\theta=\arctan(x/a)$, clean up and solve the definite integral.
A: You have the right antiderivative.  As $\theta$ approaches $\pi/2$ from below, both $\sec\theta$ and $\tan\theta$ go up to $+\infty$.  As $\theta$ approaches $-\pi/2$ from above, then $\tan\theta$ goes to $-\infty$ and $\sec\theta$ goes to $+\infty$ and when that happens, the limit can be anything, depending on which functions are involved.  One way to find that limit is to write
$$
\sec\theta+\tan\theta = \frac{1+\sin\theta}{\cos\theta}
$$
and then the numerator and denominator both go to $0$, so L'Hopital's rule can be used, and it shows the limit is $-\infty$.  Another way is via the tangent half-angle formula
$$
\sec\theta+\tan\theta = \tan\left(\frac\theta2+\frac\pi4\right).
$$
A: for $a \gt 0$
$$
\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx = 2 \int_0^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx \\
\\
\ge 2 \int_0^{\infty} \frac{1}{\sqrt{x^2+2ax+a^2}}\,dx
= 2\int_0^{\infty} \frac{1}{x+a}\,dx
$$
