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

Question

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.

Answer 1

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 $x = u^2$.

Answer 2

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.

Answer 3

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.