# Convergence of $\int^\infty_0 \frac{e^{-\sqrt x}}{1+x}dx$

I would like to know if the improper integral $$\int^\infty_0 \frac{e^{-\sqrt x}}{1+x}dx \qquad (1)$$ is convergent or not. I tried substitution and integration by parts but got no simplification. So, I wonder if someone more experienced is looking at this integral, does he immediately see the right approach (e.g., the right substitution)? Maybe there is a criterion for convergence I can apply here? I have the same problems with $$\int^\infty_1 \frac{\log(x)}{1+x}dx \qquad (2)$$ to find the right substitution. Any help is appreciated. Thanks.

For $(1)$ I think a comparison is what you want. Note that $$\frac{e^{-\sqrt{x}}}{1+x}<\frac{e^{-\sqrt{x}}}{x}< \frac{e^{-\sqrt{x}}}{\sqrt{x}}$$ for $x \geq 1$. and hence $$\int_0^\infty \frac{e^{-\sqrt{x}}}{1+x}dx = \int_0^1\frac{e^{-\sqrt{x}}}{1+x}dx+\int_1^\infty \frac{e^{-\sqrt{x}}}{1+x}dx \\ < \int_0^1\frac{e^{-\sqrt{x}}}{1+x}dx+ \int_1^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}}dx$$ where $\int_0^1\frac{e^{-\sqrt{x}}}{1+x}dx$ is clearly finite. See if you can do something with number $(2)$ along these lines.

Edited for user84413's observation.

• Excellent comparison +1 – jm324354 Jan 23 '15 at 3:28
• I just worked it out by hand and made total sense. But now, is it possible to determine the exact value this integral converges to, and not just prove it converges? – jm324354 Jan 23 '15 at 3:32
• @bd1251252 sometimes you cannot evaluate integrals exactly by hand because a closed form doesn't exist. I'd guess that is the case here. But a computer should get a good enough proximation. Fortunately the comparison test gives us an upper bound, so we might conjecture the approximate answer of the original integral :) – graydad Jan 23 '15 at 3:33
• (This is a good idea, but I think the comparison is only valid for $x>1$.) – user84413 Jan 23 '15 at 16:34
• You're right, and then this comparison works fine. – user84413 Jan 23 '15 at 16:39

Hint: Compare the integral with a integral known to converge, even one whose integrand has a closed-form antiderivative.

For example, for (1) we have $0 \leq \frac{e^{-\sqrt{x}}}{1 + x} \leq e^{-x}$ on $[1, \infty)$, so we can write $$0 \leq \int_0^{\infty} \frac{e^{-\sqrt{x}}}{1 + x} dx \leq \int_0^1 \frac{e^{-\sqrt{x}}}{1 + x} \,dx + \int_1^{\infty} e^{-x} \,dx,$$ but both of these integrals converge. Alternatively, we can observe that $0 \leq \frac{e^{-\sqrt{x}}}{1 + x} \leq e^{-\sqrt{x}}$, so $$0 \leq 0 \leq \int_0^{\infty} \frac{e^{-\sqrt{x}}}{1 + x} dx \leq \int_0^{\infty} e^{-\sqrt{x}} \,dx,$$ which we can show converges, e.g., by evaluating explicitly with the substitution $x = u^2$.

Exact value? Well, Maple does it in terms of an "exponential integral" function $$\int_0^\infty \frac{e^{-\sqrt{x}}\,dx}{1+x} = e^i \mathrm{Ei}_1(i) +e^{-i} \mathrm{Ei}_1(-i)$$ Definition: $$\mathrm{Ei}_1(z) = \int_1^\infty \frac{e^{-tz}}{t}\,dt,\qquad\mathrm{Re}(z)>0$$ and extended by analytic continuation.