Finding the derivative of $f(x)=\int_1^{\infty}\frac{e^{-xy}}{y^2}dy,\:\:\:x\in(0,\infty)$

Question

Let $$f(x)=\int_1^{\infty}\frac{e^{-xy}}{y^2}dy,\:\:\:x\in(0,\infty)$$ Show that $f(x)$ is differentiable on $(0,\infty)$, find the formula for $f'(x)$? Is $f(x)$ twice differentiable?

I'm thinking to define a sequence as follow $$f_n(x)=\int_1^n\frac{e^{-xy}}{y^2}dy.$$

To show $f_n(x)$ is differentiable, I'm thinking to show the following limit exists,

$$\lim_{h\to 0}\int_1^n\frac{e^{-(x+h)y}-e^{-xy}}{y^2h}.$$

To be able to pass the limit inside the integral, we can apply the Lebesgue dominated convergence theorem. So I want to see if I can apply it. Since $\frac{e^{-(x+h)y}-e^{-xy}}{y^2h}$ is bounded by $\frac{e^{-uy}}{y}$, (where $x\leq u\leq x+h$) which is integrable on $[1,\infty).$

Hence $f'_n(x)=\int_1^n\frac{d}{dx}(\frac{e^{-xy}}{y^2})dy=\int_1^n-\frac{e^{-xy}}{y}dy. $

Now, $\lim_{n\to\infty}\int_1^n\frac{-e^{-xy}}{y}dy=\int_{1}^{\infty}\frac{-e^{-xy}}{y}dy.$

However I see a problem here, since in fact, we have $$f'(x)=\lim_{h\to 0}\lim_{n\to{\infty}}\int_1^n\frac{e^{-(x+h)y}-e^{-xy}}{y^2h}.$$

But I'm not sure, if I'm allowed to interchange these two limits. I appreciate any hint or alternative proof.

Answer 1

Let's take two derivatives under the integral first, and we will talk about justifying it after the fact.

Consider $$f(x) = \int_1^\infty \frac{e^{-xy}}{y^2} dy.$$

Then two derivatives with respect to $x$ are easy to take here:

$$f^{\prime\prime}(x) = \int_1^\infty e^{-xy} dy = -\left.\frac{e^{-xy}}{x} \right|_{1}^\infty = \frac{e^{-x}}{x}.$$

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Note that the denominator gets completely cancelled by the chain rule. Hence, (provided we can justify the derivatives) we have $$f^{\prime}(x) = \int \frac{e^{-x}}{x} dx = e^{-x}\ln(x) + \int_1^x \ln(\tau)e^{-\tau}d\tau + C$$ for some $C \in \mathbb{R}$.

Finally, we can see that as $x \to \infty$ that $f'(x) \to 0$. We conclude that $f'(x) = e^{-x}\ln(x) + \int_1^x \ln(\tau)e^{-\tau}d\tau$.

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To justify passing the derivative through the integral, we can appeal to the measure theory version of the Leibniz integral rule.

What we need is a function that bounds $g(x,y) = \frac{d}{dx} \frac{e^{-xy}}{y^2} = -\frac{e^{-xy}}{y}$ independent of $x$ that is also integrable. For any $\delta > 0$, we have $$|g(x,y)| \le \frac{e^{-\delta y}}{y}$$ for all $x \in [\delta, \infty)$. Therefore, given $x > \delta$, we have $$\frac{d}{dx} \int_1^\infty \frac{e^{-xy}}{y^2} dy = \int_1^\infty \frac{d}{dx}\left( \frac{e^{-xy}}{y^2} \right) dy.$$ Since, $\delta$ was chosen to be an arbitrary positive number, we may conclude that this formula holds for all $x > 0$. A similar argument can be performed for the second derivative as well.

Answer 2

By Fubini's theorem,$$ f(x) = \int_1^{+\infty} \frac{\mathrm{e}^{-xy}}{y^2} \,\mathrm{d} y = \int_1^{+\infty} \int_x^{+\infty} \frac{\mathrm{e}^{-uy}}{y} \,\mathrm{d} u\,\mathrm{d} y = \int_x^{+\infty} \int_1^{+\infty} \frac{\mathrm{e}^{-uy}}{y} \,\mathrm{d} y\,\mathrm{d} u, $$ thus$$ f'(x) = -\int_1^{+\infty} \frac{\mathrm{e}^{-xy}}{y} \,\mathrm{d} y. $$