1
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Excellent comparison +1 $\endgroup$ – jm324354 Jan 23 '15 at 3:28
  • $\begingroup$ 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? $\endgroup$ – jm324354 Jan 23 '15 at 3:32
  • 1
    $\begingroup$ @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 :) $\endgroup$ – graydad Jan 23 '15 at 3:33
  • $\begingroup$ (This is a good idea, but I think the comparison is only valid for $x>1$.) $\endgroup$ – user84413 Jan 23 '15 at 16:34
  • $\begingroup$ You're right, and then this comparison works fine. $\endgroup$ – user84413 Jan 23 '15 at 16:39
1
$\begingroup$

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 $x = u^2$.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.