Regarding order of integration I have following integration problem. If i change the order of integration I get different answer. Can anybody which order of integration is correct and why?
The problem is
$$\int_0^x\int_0^\infty\frac{e^{-a/b}}{b}e^{-b}dbda $$
Now if I integrate with respect to $b$ first then 
$$\int_0^x 2K_0(2\sqrt{a})da$$
$$-2\sqrt{a}K_1(2\sqrt{a})|_0^x$$
$$-[2\sqrt{x}K_1(2\sqrt{x})-2\sqrt{0}K_1(2\sqrt{0})]$$

Now if I integrate with respect to $a$ first then the problem becomes
$$\int_0^\infty\int_0^x\frac{e^{-a/b}}{b}e^{-b}dadb$$
$$\int_0^\infty[1-e^{-x/b}]e^{-b}db$$
$$1-\int_0^\infty e^{-(x/b)-b}db$$
Now using 3.371-9 from Ryzhik and Gradshteyn i can find it to be
$$1-2\sqrt{x}K_1(2\sqrt{x})$$ 

These answer can become equal if $$2\sqrt{0}K_1(2\sqrt{0})=1$$
So can we say them to be equal?
The relevant theorem is Fubini's theorem

Here your function is non-negative, so it apply and both order gives the same result.

Now, I didn't check your calculations, so you may have a mistake somewhere in them.
The relevant theorem is Fubini's theorem

Here your function is non-negative, so it apply and both order gives the same result.

Now, I didn't check your calculations, so you may have a mistake somewhere in them.