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I used the series expansion of $e^{-x}$ and the fact that $\log(x)$ was less than $x$ in the $(0, \infty)$ to get an upper bound and so use simple comparison to show this was indeed integrable over $(0,\infty)$.

Can I do any better? Definitely can't integrate by parts or use any familiar tools but can I write in the form of a summation or something? According to Wolfram the answer is $\gamma$ the E-M constant.

Edit: Nope that's a horrific mistake: "$\log(x)$ was less than $x$"

Edit: I have realised what I said works in $(1,\infty)$. The problem is $(0,1)$ On $(1,\infty)$ $log(x) < x$ and $e^{x}=1+x+...+\frac{x^3}{6}...\geq \frac{x^3}{6}$ which is integrable.

so $e^{-x}\log(x) \leq \frac{6}{x^2} $

we integrate that in $(1,\infty)$ to get $6$.

The other region I can't figure out

Arcane1729
  • 1,188

2 Answers2

1

You should break apart the interval and isolate positive and negative contributions. The correct value is $- \gamma$ not $\gamma$.

On $[0,1]$ we have $e^{-1} \leqslant e^{-x} \leqslant 1$ and $\log x \leqslant \log x e^{-x} \leqslant e^{-1} \log x.$

Hence,

$$\int_0^1 \log x \, dx = -1 \leqslant \int_0^1 \log x e^{-x} \, dx \leqslant - e^{-1}.$$

On $[1 , \infty)$ we have using $\log x = 2 \log \sqrt{x} < 2(\sqrt{x} -1)$

$$0 < \int_1^\infty \log x e^{-x} \, dx \leqslant \int_1^\infty 2(\sqrt{x}-1) e^{-x} dx = \sqrt{\pi} \text{erfc}(1) =0.278806.$$

So a first cut (that beats $6$ and at least shows that the value is negative) is

$$-1 \leqslant \int_0^\infty \log x e^{-x} dx \leqslant -0.089073$$

Using the less tight bound $\log x < x -1$ we get $0$ as the upper bound for the integral.

RRL
  • 90,707
1

First, we write the integral of interest as

$$\int_0^\infty e^{-t}\log(t)\,dt=\int_0^x e^{-t}\log(t)\,dt+\int_x^\infty e^{-t}\log(t)\,dt \tag 1$$

Note that by integrating by parts with $u=\log(x)$ and $v=-e^{-x}$, we can write the second integral on the right-hand side of $(1)$ as

$$\int_x^\infty e^{-t}\log(t)\,dt=e^{-x}\log(x)+\int_x^\infty \frac{e^{-t}}{t}\,dt \tag 2$$

Now, in THIS ANSWER, I showed that the integral on the right-hand side of $(2)$ satisfies the bounds

$$\frac{1}{2}e^{-x}\ln\left(1+\frac{2}{x}\right)<\int_{x}^{\infty}\frac{e^{-t}}{t}dx<e^{-x}\ln\left(1+\frac{1}{x}\right)\tag 3$$

Combining $(2)$ and $(3)$ reveals

$$\frac12 e^{-x}\log(x(x+2))\le \int_x^\infty e^{-t}\log(t)\,dt\le e^{-x}\log(1+x) \tag 4$$

Letting $x=1$, we find that

$$\bbox[5px,border:2px solid #C0A000]{\frac12 e^{-1}\log(3)\le \int_1^\infty e^{-t}\log(t)\,dt\le e^{-1}\log(2)} \tag 5$$

Next, we see that the first integral on the right-hand side of $(1)$ can be written

$$\begin{align} \int_0^1 e^{-t}\log(t)\,dt&=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^\infty t^{n} e^{-t}\,dt\\\\ &=-\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!} \tag 6 \end{align}$$

We easily obtain upper and lower bounds for the rapidly convergent alternating series in $(6)$ and find that

$$\bbox[5px,border:2px solid #C0A000]{-0.796599599297056 \le \int_0^1 e^{-t}\log(t)\,dt \le -0.796599599297053} \tag 7$$

Putting together $(5)$ and $(7)$, we obtain

$$-\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!}+\frac12 e^{-1}\log(3) \le \int_0^\infty e^{-t}\log(t)\,dt \le -\sum_{n=1}^\infty \frac{(-1)^n}{n\,n!}+e^{-1}\log(2)$$

Therefore, we find that the integral of interest is bounded by

$$\bbox[5px,border:2px solid #C0A000]{-0.594521161887402 \le \int_0^\infty e^{-t}\log(t)\,dt \le -0.541605001863052} $$

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Mark Viola
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