Integral $\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx$

Question

$$\int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx$$

I subbed in $x=a \tan\theta$ and ended up with $\ln|\sec\theta+\tan\theta|$. Is this correct?

Thanks.

Answer 1

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

Answer 2

Yes, now resub $\theta=\arctan(x/a)$, clean up and solve the definite integral.

Answer 3

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 $$ \sec\theta+\tan\theta = \tan\left(\frac\theta2+\frac\pi4\right). $$

Answer 4

for $a \gt 0$ $$ \int_{-\infty}^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx = 2 \int_0^{\infty} \frac{1}{\sqrt{x^2+a^2}}\,dx \\ \\ \ge 2 \int_0^{\infty} \frac{1}{\sqrt{x^2+2ax+a^2}}\,dx = 2\int_0^{\infty} \frac{1}{x+a}\,dx $$