# $\displaystyle\int_0^\infty\frac{e^{-2x}x}{(1+e^{-x})^4} \mathrm{d}x=\frac16\left(\ln2-\frac14\right)$. But I don't know how my calculation went wrong

Let $$e^{-x}=:t$$,
get $$\def\dd{\mathrm{d}} x=-\ln t,\, \dd x = -\dfrac1t \dd t;\; t\colon1\!\backsim\!0$$,
hence \def\dd{\mathrm{d}} \begin{align*} I &= \int_1^0 \frac{t\ln t}{(1+t)^4}\dd t \\ &= \frac13\int_0^1 t\ln t\,\dd(1+t)^{-3} \\ &= \underbrace{\frac13\left.\frac{t\ln t}{(1+t)^3}\right|_0^1}_{\kern.48em=~0} +\frac13\int_1^0 (1+\ln t)(1+t)^{-3}\dd t \\ &= \frac16\int_0^1 (1+\ln t)\,\dd(1+t)^{-2} \\ &= \frac16\,\underbrace{\!\left.\frac{1+\ln t}{(1+t)^2}\right|_0^1}_{=\raise{-0.05em}{\large\frac14}-1{\color{#F03}{-\ln0}}} +\frac16\int_1^0 \frac{\dd t}{t(1+t)^2}. \end{align*}

(The part marked in $$\color{#F03}{\text{red}}$$ is exactly the strange and possibly wrong part)

As to $$\displaystyle I_1 = \int_1^0 \frac{\mathrm{d}t}{t(1+t)^2}$$,
let $$t=\dfrac1u$$,
get $$\def\dd{\mathrm{d}} u=\dfrac1t,\, \dd t = -u^{-2}\dd u;\; u\colon1\!\backsim\!\infty$$,
hence \def\dd{\mathrm{d}} \begin{align*} I_1 &= \int_\infty^1 \frac{u}{(1+u)^2}\dd u \\ &= \int_\infty^1 \frac{(1+u)-1}{(1+u)^2}\dd u \\ &= \int_\infty^1 \frac{\dd(1+u)}{1+u} - \int_\infty^1 \frac{\dd(1+u)}{(1+u)^2} \\ &= \biggl.\ln(1+u)\biggr|_\infty^1 + \left.\frac1{1+u}\right|_\infty^1 \\ &= \ln2 {\color{#F03}{-\ln(1+\infty)}} + \frac12. \end{align*}

Therefore \begin{align*} I &= \frac16\left(\frac14-1-\ln0 + \ln2-\ln(1+\infty)+\frac12\right) \\ &= \frac16\left(\ln2-\frac14\right) {\color{#F03}{-\frac16\ln(0\cdot\infty)}}. \end{align*}

(And it would be great if there was a faster way to solve it!)

• To evaluate a finite $\int_a^buv'dx=\color{blue}{[uv]_a^b}-\color{limegreen}{\int_a^bu'vdx}$ you have to ensure each coloured part is finite, which may be a simple matter of adding a suitable constant to your initial choice of $v$.
– J.G.
Aug 17, 2022 at 7:59
• @J.G. "adding a suitable constant" – how to do it specifically? (I can't figure it out)
– ooo
Aug 17, 2022 at 8:06
• Don't directly write things like $\ln(0)$ or $\ln(\infty)$. They don't make any sense .Instead see that you have $$\frac{1}{6}\lim_{h\to 0^{+}} \bigg(-\frac{\ln(h)}{(1+h)^{2}}-\ln(1+\frac{1}{h})\bigg)$$ and see that this is just $0$. Aug 17, 2022 at 8:09
• @ooo You first used IBP with $u=t\ln t,\,v=\frac13(C-(1+t)^{-3})$ where, while you chose $C=0$, I recommend $C=1$.
– J.G.
Aug 17, 2022 at 8:16
• @ooo Each of your steps were fine except for the substitution of $\infty$ and $0$. Do as I said in the comments and you'll arrive directly at the answer . $\lim_{h\to 0^{+}}\ln(1+\frac{1}{h})$ is what you get when you put infinity in $I_{1}$. Now this and the $\ln(0)$ part will cancel and the limit will be $0$ . Aug 17, 2022 at 8:58

To avoid undefined limits, integrate by parts instead as follows

\begin{align*} \int_1^0 \frac{t\ln t}{(1+t)^4}\ dt = &\ \frac16 \int_0^1\ln t\,d\left( -\frac{t^2(3+t)}{(1+t)^3}\right) \\=& \ \frac16 \int_0^1 \frac1{1+t} +\frac1{(1+t)^2} -\frac2{(1+t)^3}\ dt\\= &\ \frac16\ln2 -\frac1{24} \end{align*}

• Did you omit the IBP that is one step ahead? Otherwise, how can you "directly see" such a big lump fraction under the differential symbol?
– ooo
Aug 17, 2022 at 10:49
• @ooo - you would see it from $\int \frac{t}{(1+t)^4}\ dt = \frac{t^2(3+t)}{(1+t)^3}$ Aug 17, 2022 at 10:57
• @ooo Quanto wants you to do this :- $\frac{t}{(1+t)^{4}}=\frac{t+1}{(t+1)^{4}}-\frac{1}{(1+t)^{4}}$ and then use by Parts with the above as the integrating function and $\ln(t)$ is the differentiating one. Aug 17, 2022 at 12:01
• Ah, it turns out to be so simple, it seems that my sense of mathematics is too bad, a little embarrassing, huh huh. Thank you all, especially @Mr.GandalfSauron for bothering to annotate such basic stuff for me. I am grateful to every enthusiastic netizen in this community.
– ooo
Aug 17, 2022 at 13:48
• $\displaystyle +1$. Fine. Aug 17, 2022 at 20:05

As in the comments Don't directly write things like $$\ln(0)$$ or $$\ln(∞)$$. They don't make any sense .Instead see that you have $$\frac{1}{6}\lim_{h\to 0^{+}} \bigg(-\frac{\ln(h)}{(1+h)^{2}}-\ln(1+\frac{1}{h})\bigg)$$ and this is just $$0$$. Note that you get $$\lim_{h\to 0^{+}}\ln(1+\frac{1}{h})$$ due to your substitution of $$\infty$$ in $$I_{1}$$. I have taken the limit to $$0$$ for easier computation. And your $$-\frac{\ln(h)}{(1+h)^2}$$ comes from your substitution of $$0$$ in the first integral. So when the limit is evaluated, you end up with $$0$$ and you arrive at the desired answer.

Another way to do it is to expand into a series and then use Dominated Convergence Theorem.

$$\frac{-1}{6}\int_{0}^{\infty}\sum_{r=3}^{\infty}(-1)^{r}r(r-1)(r-2)e^{-(r-3)x}e^{-2x}x\,dx$$ $$\frac{-1}{6}\int_{0}^{\infty}\sum_{r=3}^{\infty}(-1)^{r}r(r-1)(r-2)e^{-(r-1)x}x\,dx$$

$$=\frac{-1}{6}\sum_{r=3}^{\infty}(-1)^{r}\frac{r(r-1)(r-2)}{(r-1)^{2}}$$

$$=\frac{-1}{6}\sum_{k=2}^{\infty}(-1)^{k}\frac{(k+1)(k-1)}{k}$$

Now $$\ln(1+x)=\sum_{r=1}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}=x+\sum_{r=2}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}$$

So $$\ln(1+x)-x=\sum_{r=2}^{\infty}\frac{(-1)^{r-1}x^{r}}{r}$$

Now multiply by $$x$$ on both sides and differentiate to get

$$\frac{x}{1+x}+\ln(1+x)-2x=\sum_{r=2}^{\infty}(-1)^{r}(r+1)\frac{x^{r}}{r}$$

Now divide by $$x$$ on both sides and differentiate again to get

$$-\frac{1}{(1+x)^{2}}+\frac{1}{x(1+x)}-\frac{-\ln(1+x)}{x^{2}}=\sum_{r=2}^{\infty}(-1)^{r}(r+1)(r-1)\frac{x^{r-1}}{r}$$ .

Now put $$x=1$$

PS:- There could be some errors with the sign in some steps , but I have shown you the idea to compute the integral. I hope you'll be able to compute it and finish it on your own now.

– ooo
Aug 17, 2022 at 9:07
• I have edited it in Aug 17, 2022 at 9:12

The problem is the line that says

$$"= \frac16\,\underbrace{\!\left.\frac{1+\ln t}{(1+t)^2}\right|_0^1}_{=\raise{-0.05em}{\large\frac14}-1{\color{#F03}{-\ln0}}} +\frac16\int_1^0 \frac{dt}{t(1+t)^2},"$$

because directly plugging in 0 results in an undefined term. In general, the Integration by Parts Theorem says that if $$f,g$$ are differentiable on $$\left[a,b\right]$$ with $$f',g'$$ integrable on $$\left[a,b\right]$$, then

$$\int_{a}^{b}f'\left(x\right)g\left(x\right)dx\ =\ f\left(b\right)g\left(b\right)-f\left(a\right)g\left(a\right)-\int_{a}^{a}f\left(x\right)g'\left(x\right)dx.$$

Basically, we need each of those three terms to the right of the equal sign to be finite.

But to finish the question, let's take the limit as $$\epsilon \to 0$$ (when $$\epsilon > 0$$) of the original integral. More specifically, let's solve

$$\lim_{\epsilon \to 0}\frac{1}{3}\int_{1}^{\epsilon}\left(1+\ln\left(t\right)\right)\left(1+t\right)^{-3}dt.$$

(Note that your work is still correct even before this integral.)

Using your IBP choice, we get the integral to equal

$$-\frac{1}{3}\lim_{\epsilon \to 0}\left(-\frac{1}{2}\frac{1+\ln{(t)}}{(1+t)^2}\Bigg|_\epsilon^1 - \int_{\epsilon}^1-\frac{dt}{2t(t+1)^2}\right).$$

Notice that we aren't distributing the limit symbol to the two terms because we can only do that with finite terms. After all,

$$\lim_{\epsilon \to 0} -\frac{1}{2}\left(\frac{1+\ln\left(\epsilon\right)}{\left(1+\epsilon\right)^{2}}\right) = \infty.$$

So we shall save computing the limit until the end when we finished evaluating and simplifying an antiderivative. We can use partial fractions on the second integral and do some grunt work. Eventually, we'll get the original integral to be

\eqalign{ -\lim_{\epsilon \to 0}\left[\frac{x\left(x+2\right)\ln\left(x\right)+x}{6\left(x+1\right)^{2}}-\frac{\ln\left(x+1\right)}{6}\right]_{\epsilon}^{1} &= -\frac{1}{24}+\frac{\ln\left(2\right)}{6} + \lim_{\epsilon \to 0}\frac{\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon}{6\left(\epsilon+1\right)^{2}}. }

To show that limit converges to $$0$$, we can do

\eqalign{ \lim_{\epsilon \to 0}\frac{\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon}{6\left(\epsilon+1\right)^{2}} &= \frac{\lim_{\epsilon \to 0}\left(\epsilon\left(\epsilon+2\right)\ln\left(\epsilon\right)+\epsilon\right)}{\lim_{\epsilon \to 0}6\left(\epsilon+1\right)^{2}}. }

It's a little more work to use L'Hôpital's Rule on the numerator, but it goes to $$0$$.

Does that help?

• "some grunt work"... indeed. Thanks for your patient typing, but except for pointing out the caveat stating "convergence required", IMHO the rest doesn't seem to be of much help, but thanks anyway, hard work!
– ooo
Aug 17, 2022 at 10:35