I would like to calculate $$ F(\alpha)=\int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx $$ for $\alpha>0.$

Since WolframAlpha provides a complicated expression in terms of hypergeometric functions and the exponential integral $\mathrm{Ei}$ function already for $\alpha=1$, I see little hope for an exact analytical evaluation. Therefore I tried to estimate $F(\alpha)$ at least in some limit, as follows: integrating repeatedly by parts, $$\begin{aligned} f(\alpha)\equiv -F'(\alpha)&=\int_0^\infty \log(1+x)e^{-\alpha x}dx\\ &=-\frac{1}{\alpha}\log(1+x)e^{-\alpha x}\bigg|_0^\infty+\frac{1}{\alpha}\int_0^\infty \frac{e^{-\alpha x}}{1+x}dx\\ &=0-\frac{e^{-\alpha x}}{\alpha^2(1+x)}\bigg|_0^\infty-\frac{1}{\alpha^2}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^2}dx\\ &=\frac{1}{\alpha^2}+\frac{e^{-\alpha x}}{\alpha^3(1+x)^2}\bigg|_0^\infty+\frac{2}{\alpha^3}\int_0^\infty \frac{e^{-\alpha x}}{(1+x)^3}dx\\ &=\frac{1}{\alpha^2}-\frac{1}{\alpha^3}-\frac{2e^{-\alpha x}}{\alpha^4(1+x)^3}\bigg|_0^\infty-\frac{3!}{\alpha^4}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^4}dx. \end{aligned}$$ Proceeding by induction, $$ f(\alpha)=\frac{1}{\alpha^2}\left(1-\frac{1}{\alpha}+\frac{2}{\alpha^2}-\ldots+\frac{(-1)^nn!}{\alpha^n}\right)+(-1)^{n+1}\frac{(n+1)!}{\alpha^{n+2}}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^{n+2}}dx, $$ from which $$\begin{aligned} \lim_{\alpha\to\infty}\alpha^{n+2}\left[f(\alpha)-\sum_{j=0}^{n-1}\frac{(-1)^jj!}{\alpha^{j+2}}\right]&=(-1)^{n}{n!}+(-1)^{n+1}(n+1)!\lim_{\alpha\to\infty}\int_0^\infty\frac{e^{-\alpha x}}{(1+x)^{n+2}}dx\\ &=(-1)^{n}{n!} \end{aligned}$$ by dominated convergence. This shows that we have obtained an asymptotic series for $f(\alpha)$ in the limit $\alpha\to\infty$: $$ f(\alpha)\underset{\alpha\to\infty}{\sim}\sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\ . $$ Even if it is of course divergent, this series is Borel summable: $$\boxed{ \sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\overset{B}{=}\int_0^\infty dt\, e^{-t}\sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\frac{t^n}{n!}\,{\color{red}=}\,\frac{1}{\alpha^2}\int_0^\infty e^{-t(1+1/\alpha)}dt=\frac{1}{\alpha(1+\alpha)}}\ . $$ It is therefore tempting to state that $$ f(\alpha)=\int_0^\infty \log(1+x)e^{-\alpha x}dx\overset{?}{=}\frac{1}{\alpha(1+\alpha)} $$ and hence, integrating in $\alpha$ and setting the integration constant to zero because $F(\alpha)\to0$ as $\alpha\to\infty$ by dominated convergence, $$ F(\alpha)=\int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx \overset{?}{=} \log\frac{1+\alpha}{\alpha}. $$ Strictly speaking, these equalities are wrong, but it appears that they provide reliable numerical approximations to $f(\alpha)$ and $F(\alpha)$ in the $\alpha\to\infty$ limit.

How do I keep track of the degree of approximation involved in the resummation? This was explicit in the asymptotic series, which is however not as nice to write down.

EDIT: Maybe it can be instructive to compare with a similar situation where the exact solution is available. Consider $$ G(\alpha)=\int_0^\infty\frac{\sin x}{x}e^{-\alpha x}dx $$ for $\alpha>0$. Here $$ g(\alpha)\equiv-G'(\alpha)=\int_0^\infty \sin x\, e^{-\alpha x}dx=\mathrm{Im}\left[\int_0^\infty e^{-(\alpha-i)x}dx\right]=\mathrm{Im}\left[\frac{1}{\alpha-i}\right]=\frac{1}{1+\alpha^2} $$ and hence, because $G(\alpha)\to0$ as $\alpha\to\infty$, $$ G(\alpha)=\frac{\pi}{2}-\arctan \alpha. $$ Nevertheless we can work out an asymptotic series for $g(\alpha)$ by integrating by parts. The result is $$ g(\alpha)\underset{\alpha\to\infty}{\sim}\sum_{n=0}^\infty \frac{(-1)^n}{\alpha^{2n+2}}. $$ In this case however the series converges to the exact result $g(\alpha)=1/(1+\alpha^2)$ for $\alpha>1$.

EDIT 2: as was pointed out by Sangchul Lee in the comments, the derivation in the box contains a mistake $$ \sum_{n=0}^{^\infty}\frac{(-1)^n n!}{\alpha^{n+2}}\overset{B}{=}\int_0^\infty \frac{e^{-t}}{\alpha(t+\alpha)}dt\,{\color{red}\neq}\,\frac{1}{\alpha(1+\alpha)}, $$ which is one of the integral representations of $f(\alpha)$ already appearing in the second equation. Therefore, Borel resummation gives the exact result. What I still don't know is how the wrong expressions I derived give a reasonable approximation for large $\alpha$.

  • $\begingroup$ I would conjecture that $f(a) = \frac{-e^a \operatorname{Ei}(-a)}{a}$, at least for integer $a$. $\endgroup$ – Brevan Ellefsen Apr 9 '17 at 19:09
  • $\begingroup$ very interesting question. i also encountered this phenomena and it occured that it yielded an exact closed form time to time (search my answer here on mse, maybe you'll find something) which is really fascinating $\endgroup$ – tired Apr 9 '17 at 19:25
  • $\begingroup$ @BrevanEllefsen yepp, also for non integer $a$ but reintegrating gives us pure evil: a meijer-g occurs $\endgroup$ – tired Apr 9 '17 at 19:26
  • 1
    $\begingroup$ In other direction, it seems that we have the following asymptotic expansion $$ F(\alpha) = \frac{1}{2}\log^2\alpha + \gamma \log\alpha + \frac{1}{2}(\gamma^2 + 3\zeta(2)) + o(1) $$ as $\alpha \to 0^+$. $\endgroup$ – Sangchul Lee Apr 12 '17 at 4:29
  • 1
    $\begingroup$ Also, in your computation of the Borel summation, you should have $$ \sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}\frac{t^n}{n!} = \frac{1}{\alpha(t + \alpha)}, $$ which indeed yields the integral representation of $f(\alpha)$. $\endgroup$ – Sangchul Lee Apr 12 '17 at 5:01

I'll provide a somehow different approach to approximate the integral

$$F(\alpha) = \int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx$$

Using integration by parts

$$F(\alpha) = \int_0^\infty \frac{\log(1+x)}{x}e^{-\alpha x}dx=-\alpha\int^\infty_0 e^{-\alpha x}\mathrm{Li}_2(-x)\,dx$$

For the record this integral appears in Lewis book

$$\int^\infty_0 e^{-\alpha x} \mathrm {Li}_2 (-x) \, dx = \frac {1}{\alpha}\int^\infty_{\alpha}\frac {e^x}{x}\mathrm {Ei}(-x)\, dx$$

Hence we have

$$F(\alpha) = \int^\infty_{\alpha}\frac {e^x}{x}\mathrm {E}_1(x)\, dx$$

Now use the approximation

$$\frac{1}{2}e^{-x}\log\left( 1+\frac{2}{x}\right)<\mathrm{E}_1(x) < e^{-x}\log\left( 1+\frac{1}{x}\right)$$

Wiki picture showing the tightness of this bound

enter image description here

Hence we have

$$\frac{1}{2} \int^\infty_{\alpha}\frac{\log\left( 1+\frac{2}{x}\right)}{x}\,dx<F(\alpha) < \int^\infty_{\alpha}\frac{\log\left( 1+\frac{1}{x}\right)}{x}\,dx$$

This can be rewritten as

$$ -\frac{1}{2}\mathrm{Li}_2\left( -\frac{2}{\alpha}\right)< F(\alpha) < -\mathrm{Li}_2\left( -\frac{1}{\alpha}\right)$$

A plot for 30 points for $\alpha$ showing the upper and lower bounds

enter image description here

A plot for 10 points for $\alpha$ showing the upper and lower bounds

enter image description here

A scatter plot for 10 points

enter image description here


Behavior of the Laplace transform $\mathcal{L}\{g\}(s)$ for large parameter $s$ is intimately related to the near-zero behavior of the function $g$ being transformed. In particular, the leading order of $\mathcal{L}\{g\}(s)$ as $s \to \infty$ is very robust and roughly depends only on the value $g(0)$. That is a reason why your computation still gives a good approximation.

Next, let me derive an asymptotic expansion for $F(\alpha)$. We begin by splitting the integral into two parts

$$F(\alpha) = \int_{0}^{\epsilon} \frac{\log(1+x)}{x} e^{-\alpha x} \, dx + \int_{\epsilon}^{\infty} \frac{\log(1+x)}{x} e^{-\alpha x} \, dx. $$

Applying the Cauchy-Schwarz inequality to the latter term, we have

\begin{align*} \left|\int_{\epsilon}^{\infty} \frac{\log(1+x)}{x} e^{-\alpha x} \, dx\right| &\leq \bigg( \int_{\epsilon}^{\infty} \frac{\log^2(1+x)}{x^2}\, dx \bigg)^{1/2}\bigg( \int_{\epsilon}^{\infty} e^{-2\alpha x} \, dx \bigg)^{1/2} \\ &\leq C\alpha^{-1/2}e^{-\epsilon\alpha} \end{align*}

and thus the latter term only contributes to an exponentially decaying error. On the other hand, if $\epsilon < 1$ and $N$ is a positive integer, then

\begin{align*} \int_{0}^{\epsilon} \frac{\log(1+x)}{x} e^{-\alpha x} \, dx &= \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n} \int_{0}^{\epsilon} x^{n-1} e^{-\alpha x} \, dx \\ &\qquad + \int_{0}^{\epsilon} \bigg( \frac{\log(1+x)}{x} - \sum_{n=1}^{N} \frac{(-1)^{n-1}}{n}x^{n-1} \bigg) e^{-\alpha x} \, dx \\ &= \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \frac{1}{\alpha^n} \int_{0}^{\alpha \epsilon} x^{n-1} e^{-x} \, dx \\ &\qquad + \mathcal{O}\bigg( \int_{0}^{\infty} x^N e^{-\alpha x} \, dx \bigg). \end{align*}

Here, for each fixed $n$, it is not hard to check that

$$\int_{0}^{\alpha \epsilon} x^{n-1} e^{-x} \, dx = (n-1)! + \mathcal{O}\big( (\epsilon\alpha)^{n-1}e^{-\epsilon\alpha} \big). $$

Putting altogether, for each fixed $N$ we have an asymptotic expansion

$$ F(\alpha) = \sum_{n=0}^{N-1} \frac{(-1)^n n!}{n+1} \frac{1}{\alpha^{n+1}} + \mathcal{O}\bigg(\frac{1}{\alpha^{N+1}}\bigg) \qquad \text{as } \alpha \to \infty $$

It is not surprising that the first term is also obtained formally by integrating OP's asymptotic expansion for $f(\alpha)$ as well.

For the near-zero behavior of $F$, we begin from the expression @Zaid Alyafeai derived:

$$ F(\alpha) = -\alpha \int_{0}^{\infty} \operatorname{Li}_2(-x) e^{-\alpha x} \, dx = -\int_{0}^{\infty} \operatorname{Li}_2(-x/\alpha) e^{-x} \, dx. $$

Utilizing the identity

$$ -\operatorname{Li}_2(-z) = \zeta(2) + \frac{1}{2}\log^2 z + \operatorname{Li}_2 (-1/z), $$

we find that

\begin{align*} F(\alpha) &= \int_{0}^{\infty} \left( \zeta(2) + \frac{1}{2}\log^2(x/\alpha) + \operatorname{Li}_2(-\alpha/x) \right) e^{-x} \, dx \\ &= \frac{1}{2}\log^2\alpha + \gamma \log\alpha + \frac{1}{2}(\gamma^2 + 3\zeta(2)) + \int_{0}^{\infty} \operatorname{Li}_2(-\alpha/x) e^{-x} \, dx. \end{align*}

We claim that the last integral vanishes as $\alpha \to 0$. Indeed, from the estimate $|\operatorname{Li}_2(-x^{-1})| \sim x^{-1}$ as $x \to \infty$, it is possible to prove that

$$ \int_{0}^{\infty} \operatorname{Li}_2(-\alpha/x) e^{-x} \, dx = \alpha \int_{0}^{\infty} \operatorname{Li}_2(-1/x) e^{-\alpha x} \, dx = \mathcal{O}\big(\alpha \log (1/\alpha) \big). $$


$$F(\alpha) = \frac{1}{2}\log^2\alpha + \gamma \log\alpha + \frac{1}{2}(\gamma^2 + 3\zeta(2)) + \mathcal{O}\big(\alpha \log (1/\alpha) \big) \qquad \text{as } \alpha \to 0^+. $$

In fact a more detailed analysis on the error term is available. Let

$$ g(\alpha) = \int_{0}^{\infty} \operatorname{Li}_2(-\alpha/x) e^{-x} \, dx. $$

Then $g$ defines a continuous function on $[0,\infty)$ which is differentiable on $(0,\infty)$. Differentiating under the integral sign,

$$ g'(\alpha) = - \int_{0}^{\infty} \frac{1-e^{-x}}{x(x+\alpha)}\, dx = - \frac{1}{\alpha} \int_{0}^{\infty} \left( \frac{1}{x} - \frac{1}{x+\alpha} \right)(1-e^{-x}) \, dx. $$

The last integral exhibits cancellation of singularity at infinity. In order to analyze this effect, we work with the truncated integral.

\begin{align*} &\int_{0}^{R} \left( \frac{1}{x} - \frac{1}{x+\alpha} \right)(1-e^{-x}) \, dx \\ &= \int_{0}^{\alpha} \frac{1-e^{-x}}{x} \, dx + \int_{\alpha}^R \frac{1-e^{-x}}{x} \, dx - \int_{\alpha}^{R+\alpha} \frac{1-e^{-(x-\alpha)}}{x} \, dx \\ &= \int_{0}^{\alpha} \frac{1-e^{-x}}{x} \, dx + \log \left(\frac{R}{R+\alpha}\right) - \int_{\alpha}^R \frac{e^{-x}}{x} \, dx + e^{\alpha}\int_{\alpha}^{R+\alpha} \frac{e^{-x}}{x} \, dx. \end{align*}

Taking limit as $R\to\infty$, we find that

$$ g'(\alpha) = - \frac{1}{\alpha} \int_{0}^{\alpha} \frac{1-e^{-x}}{x} \, dx - \frac{e^{\alpha} - 1}{\alpha} E_1(\alpha), \tag{1} $$

where $E_1(\alpha) = \int_{\alpha}^{\infty} e^{-x}/x \, dx$ is a variant of the exponential integral. It is well-known that $E_1(\alpha)$ has the following expansion

$$ E_1(\alpha) = -\gamma - \log \alpha - \sum_{n=1}^{\infty} \frac{(-1)^n}{n!n} \alpha^n $$

which is valid for all $\alpha \in \Bbb{C} \setminus (-\infty, 0]$. Plugging this back, $g'(\alpha)$ take the form

$$ g'(\alpha) = \frac{e^{\alpha} - 1}{\alpha} \log \alpha + \text{[entire function in $\alpha$]}. $$

Mathematical tells that this entire-function part has a neat series expansion, yielding

$$ g'(\alpha) = \frac{e^{\alpha} - 1}{\alpha} \log \alpha - \sum_{n=1}^{\infty} \frac{-\gamma + H_n}{n!}x^{n-1}. $$

I haven't checked this by myself, but I expect that this is not hard to verify by expanding everything in $\text{(1)}$. Integrating, end up with the following series expansion

$$F(\alpha) = \frac{1}{2}\log^2\alpha + \gamma \log\alpha + \frac{1}{2}(\gamma^2 + 3\zeta(2)) + \sum_{n=1}^{\infty} \frac{\alpha^n}{n!n} \left(\gamma + \log \alpha - H_n - \frac{1}{n} \right). $$

For instance, the following is a numerical computation of $F(2)$ using both numerical integration and the formula above:

enter image description here

I suspect that more systematic approach is available for both directions, but I am not sure how to proceed.


Elaborating slightly on the answer by Sangchul Lee, I would like to elucidate why my (wrong) computation still gives a reasonable approximation. The mistake was made in writing $$ f(\alpha)=\int_0^\infty \frac{e^{-\alpha t}}{\alpha(t+1)}dt\,{\color{\red}=}\,\frac{1}{\alpha(1+\alpha)}. $$ Since $e^{-\alpha t}$ attains its global maximum on the integration region when $t=0$, this integral may be approximated for large $\alpha$ by Laplace's method: $$ f_\epsilon(\alpha) = \frac{1}{\alpha}\sum_{n=0}^\infty (-1)^n\int_0^{\epsilon}t^n e^{-\alpha t}dt; $$ replacing $\epsilon$ with $\infty$ now only introduces exponentially small errors so that $$ f(\alpha)\underset{\alpha\to\infty}\sim \sum_{n=0}^\infty \frac{(-1)^nn!}{\alpha^{n+2}}=\frac{1}{\alpha^2}-\frac{1}{\alpha^3}+\mathcal O(\alpha^{-4}). $$ By chance, for large $\alpha$, $$ \frac{1}{\alpha(1+\alpha)}=\frac{1}{\alpha^2}\frac{1}{1+1/\alpha}=\frac{1}{\alpha^2}\left(1-\frac{1}{\alpha}+\mathcal O(\alpha^{-2})\right)=\frac{1}{\alpha^2}-\frac{1}{\alpha^3}+\mathcal O(\alpha^{-4}). $$ So the mistake done in the computation gives the right answer up to order $\alpha^{-4}$ because it is the sum of a geometric series which agrees with the right asymptotic expansion for $f(\alpha)$ precisely up to that order.


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