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While I was looking for various methods to evaluate an indefinite integral, I met the strange equality below: $$\int_{-1}^0x\ln(1+x)dx=3\int_{0}^1x\ln(1+x)dx$$ which is equal to $\frac34$. We can verify the equality by evaluating the definite integrals in the equation by directly finding the antiderivative of the integrand or by series method, or some other way.

My question is maybe not as interesting as this equality: Can we show this equality without evaluating the two indefinite integrals, for example by a transformation from one side to the other side? I don't have any idea.

All ideas are greatly appreciated.

Bob Dobbs
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    Rephrasing the question, without explicit evaluation, you want to show that$$\int_0^1x\log\left[(1-x)(1+x)^3\right],dx=0$$Maybe there's an argument to be made that leverages symmetry of the integrand about the real root of $(1-x)(1+x)^3-1$... – user170231 Nov 11 '24 at 16:02
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    From here, there's also: $$\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right) \ln\left((1-x)(1+x)^3\right)}{x}dx=0$$ – Zacky Nov 11 '24 at 18:52

3 Answers3

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It sounds like the only rule is that we must avoid explicitly integrating $x\log(1+x)$. If we're equipped with

$$\frac d{dx}\log(1\pm x)=\frac1{x\pm1}$$

then we can convert the logs to definite integrals and apply Fubini's theorem:

$$\begin{align*} I_1 &= \int_{-1}^0 x \log(1+x) \, dx \\ &= - \int_0^1 x \log(1-x) \, dx & x\to-x \\ &= \int_0^1 \int_0^x \cdots \, dy \, dx = \int_0^1 \int_y^1 \frac x{1-y} \, dx \, dy \\ &= \frac12 \int_0^1 (1+y) \, dy \\[2ex] I_2 &= 3 \int_0^1 x \log(1+x) \, dx \\ &= \int_0^1 \int_0^x \cdots \, dy \, dx = \int_0^1 \int_y^1 \frac{3x}{1+y} \, dx \, dy \\ &= \frac32 \int_0^1 (1-y) \, dy \end{align*}$$

By symmetry about $y=\dfrac12$,

$$I_2 - I_1 = \int_0^1 (1-2y)\,dy = 0$$

user170231
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Note that the following integral vanishes

\begin{align} \int_{0}^1x\ln[(1+x)^3(1-x)]dx=& \ \frac12\int_{0}^1\ln[(1+x)^3(1-x)]d(x^2-1)\\ \overset{ibp}=&\int_0^1 (2x-1)dx=0 \end{align}

Quanto
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We can show that

$$\int_{-1}^{0}x\ln(1+x)dx = -\int_{0}^{1}x\ln(1-x)dx$$

So the limits on the RHS and LHS match the limits in your question. The hard part is finding how

$$-\int_{0}^{1}x\ln(1-x)dx = 3\int_{0}^{1}x\ln(1+x)dx$$

I don't know if this is possible without some sort of evaluation of the integral.

Kendall
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  • I did not downvote you. But, perhaps, you should had written your idea as a comment. I am sorry that you got a downvote. – Bob Dobbs Nov 11 '24 at 18:24