Let $K_a$ be the modified Bessel function of the second kind of order $a \geq 0$:
$$K_a(x)=\int^{\infty}_0\exp(-x\cosh(t))\cosh(at)dt$$
$x\in(0,\infty)$
Fix $a>0$ and use the comparison test to analyse the convergence of the integral
Well here's my attempt:
I'm going to try to expand with $\cosh(t)=\frac{e^t+e^{-t}}{2}$ to see whether I can rationalize and compare the integral to something that I know converges:
$$\int^{\infty}_0e^{-x\left(\frac{e^t+e^{-t}}{2}\right)}\frac{e^{at}+e^{-at}}{2}dt=\frac{1}{2}\int^{\infty}_0e^{-x\left(\frac{e^t+e^{-t}}{2}\right)+at}+e^{-x\left(\frac{e^t+e^{-t}}{2}\right)-at}dt=\frac{1}{2}\int^{\infty}_0e^{-x\left(\frac{e^t+e^{-t}}{2}\right)+at}dt+\frac{1}{2}\int^{\infty}_0e^{-x\left(\frac{e^t+e^{-t}}{2}\right)-at}dt$$
I had an idea to compare the integrals seperately, for example i'd like to compare them to $\int^{\infty}_0e^{-x}dx$ which we know converges. But then I have a difficulty showing that $-x\left(\frac{e^t+e^{-t}}{2}\right)+at>-x$ in order for the inequality $\frac{1}{e^{-x\left(\frac{e^t+e^{-t}}{2}\right)+at}}<\frac{1}{e^{-x}}$ to hold . Am I at all going the right direction. Any help would be wonderful
You can start performing the change $e^t=y$
$$K_a(x)=\int^{\infty}_0\,e^{-x\cosh(t)}\cosh(at)dt=\frac{1}{2}\int_1^{\infty}e^{-\frac{x}{2}(y+1/y)}(y^{a-1}+y^{-a-1})dy$$
Now, taking account $-\frac{x}{2}(y+1/y)<-\frac{x}{2}y$ and $y^{-a-1}<1$ for any $y\in(1,\infty)$ you get
$$K_a(x)\leq\frac{1}{2}\int_1^{\infty}e^{-\frac{x}{2}y}\,y^{a-1}dy+\frac{1}{2}\int_1^{\infty}e^{-\frac{x}{2}y}\,dy<\boxed{\frac{1}{2}\left(\frac{2}{x}\right)^a\Gamma(a,x/2)+\frac{1}{x}\Gamma(1,x/2)}$$ So both incomplete Gamma converges, your integral too.