Question

$\displaystyle \bf \large\int\limits^1_0 \dfrac{log (x + 1)}{x} dx$
​

Answer 1

Answer:

0

∫

1

x

log(x+1)

dx

To proceed further with this, substitute u=-x.

$-{\displaystyle\int} \sf -\dfrac{\ln\left(1-u\right)}{u}\, du−∫− </p><p>u$

ln(1−u)

du ----(1)

This is the Spence's function!

It is a special integral!

${Li _{2}(z)=\displaystyle \sf -\int _{0}^{z}{\ln(1-u) \over u}\,du{\text{, }}z\in \mathbb {C} }Li </p><p>2</p><p> </p><p> (z)=−∫ </p><p>0</p><p>z$

u

ln(1−u)

du, z∈C

So, we can rewrite equation (1). We get:

$</p><p></p><p>\sf -Li_2\left(u\right)−Li </p><p>2$

(u)

After substituting u = -x, we get:

$\sf -Li_2\left(-x\right)+constant−Li </p><p>2$

(−x)+constant

⊱ ────── ✯ ────── ⊰

Answer 2

If you can allow a bit of error, I hope approximating ln(1-x) to -x -x^2/2 and ln(1+x) to x + x^2/2…

We finally have, -x^2(1+x/2)^3….

-x^2(1+(x^6)/8+3(x^2)/2+3(x^4)/4)

=- [(x^2)/2 + (x^8)/8 + 3(x^4)/2+ 3(x^6)/4]

= -[1/6+1/72+3/10+3/28]

= -.587698

And the actual answer was obtained by one of the members here which is -0.4058

Expanding up to three terms gives you even better answer.

_/\_ Approximations _/\_

$Although this integral has already been excellently solved long ago by Aadit Pandey, I thought I might add another potential route to solve the integral, if only to explore interesting new identities.</p><p></p><p>One way to attempt to solve the integral is to use the series expansion of</p><p></p><p>log2(1+x)log2⁡(1+x)and log(1−x).log⁡(1−x).</p><p></p><p>Specifically, where HkHk denotes the k-th harmonic number,</p><p></p><p>Hk=∑n=1k1nHk=∑n=1k1n</p><p></p><p>log2(1+x)=2∑n=2∞Hn−1(−x)nnlog2⁡(1+x)=2∑n=2∞Hn−1(−x)nn</p><p></p><p>Which can be obtained by integrating the series expansion of</p><p></p><p>log(1−x)1−x=−∑n=1∞Hnxnlog⁡(1−x)1−x=−∑n=1∞Hnxn</p><p></p><p>and substituting in −x.−x.</p><p></p><p>Thus the integral, accepting the interchange of sum and integral, becomes,</p><p></p><p>I=2∑n=2∞Hn−1(−1)nn∫10xn−1log(1−x)dxI=2∑n=2∞Hn−1(−1)nn∫01xn−1log⁡(1−x)dx</p><p></p><p>Next, we seek to find an expression for</p><p></p><p>Jn=∫10xnlog(1−x)dxJn=∫01xnlog⁡(1−x)dx</p><p></p><p>There are several ways to attempt to compute this integral. Perhaps the easiest is with the integral definition of the logarithmic function itself:</p><p></p><p>log(1−x)=−∫x011−ydylog⁡(1−x)=−∫0x11−ydy</p><p></p><p>Jn=−∫10∫x0xn1−ydydxJn=−∫01∫0xxn1−ydydx</p><p></p><p>Now, visualizing the region 0<x<1,0<y<x0<x<1,0<y<x , it becomes apparent that this is identical to the region 0<y<1,y<x<10<y<1,y<x<1. Thus our double integral, with a reversal of the order of integration, is equal to,</p><p></p><p>Jn=−∫10∫1yxn1−ydxdyJn=−∫01∫y1xn1−ydxdy</p><p></p><p>Evaluating the inner integral gives,</p><p></p><p>Jn=−1n+1∫101−yn+11−ydyJn=−1n+1∫011−yn+11−ydy</p><p></p><p>But since the inside function is a simple (finite) geometric sequence,</p><p></p><p>1−yn+11−y=1+y+y2+...+yn1−yn+11−y=1+y+y2+...+yn</p><p></p><p>It is not hard to see that the integral simply becomes</p><p></p><p>Jn=−1n+1(1+12+13+...+1n+1)=−Hn+1n+1Jn=−1n+1(1+12+13+...+1n+1)=−Hn+1n+1</p><p></p><p>So we have, substituting this result into our original integral</p><p></p><p>∫10xn−1log(1−x)dx=Jn−1=−Hnn∫01xn−1log⁡(1−x)dx=Jn−1=−Hnn</p><p></p><p>We find that</p><p></p><p>I=−2∑n=2∞(−1)nHnHn−1n2I=−2∑n=2∞(−1)nHnHn−1n2</p><p></p><p>From here, though, I am not entirely sure if this can easily be shown to be what the result was shown to be in other answers, namely</p><p></p><p>I=−π4240I=−π4240</p><p></p><p>I am tempted to believe that a method should exist based on famous identities involving harmonic numbers and fourth powers of ππ, such as</p><p></p><p>∑n=1∞H2nn2=17π4360∑n=1∞Hn2n2=17π4360</p><p></p><p>∑n=1∞Hnn3=π472∑n=1∞Hnn3=π472</p><p></p><p>Notably, though, almost all of these identities involving the harmonic numbers do not possess the alternating nature that the sum obtained here does, putting a significant damper on making any easy progress.</p><p></p><p>Still, regardless of whether a method exists to evaluate this sum or not, perhaps this integral and similar logarithmic ones, with the help of the technique outlined by Aadit Pandey can serve as a ‘bridge’ to help to begin to evaluate some of these alternating sums involving harmonic numbers.</p><p></p><p>$