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I was trying to compute the integral given in this question using Laplace transform.

I began by declaring

$$\ I(a) = \int_0^{1} \frac{x^a-1}{\operatorname{ln}x\cdot (1+x^3)} dx$$

Taking the Laplace transform on both sides, we have:

$$\mathcal L(I(a))(s) = \int_0^{1}\frac{ \mathcal L(x^a) - \mathcal L(1)}{\operatorname{ln}x\cdot (1+x^3)} dx$$ $$ = \int_0^{1} \frac{dx}{s\cdot (s-\operatorname{ln}x) \cdot (1+x^3)}$$

Then I substituted $s-\operatorname{ln}x = t$. The Laplace transform then simplifies into

$$\mathcal L(I(a)(s) = \frac{e^s}{s} \cdot \int_s^{\infty} \frac{e^{-t}}{t(1+e^{3(s-t)})}dt$$

$$\mathcal L(I(a)(s) = \frac{e^s}{s} \cdot \int_s^{\infty} \frac{e^{-t}}{t}\cdot \sum_{n=0}^{\infty} \big(e^{(2n)(3s-3t)} - e^{(2n+1)(3s-3t)}\big)dt$$ $$ = \frac{1}{s} \cdot \sum_{n=0}^{\infty} \int_s^{\infty} \frac{e^{(6n+1)(s-t)} - e^{(6n+4)(s-t)}}{t} dt $$

I am unable to proceed with this. Can anyone help me evaluate this expression?

Any help is appreciated. Thank you for reading.

  • 1
    Using Mathematica ,for: $a\geq 0$ we have:$\int_0^1 \frac{x^a-1}{\ln (x) \left(1+x^3\right)} , dx=\ln \left(\frac{\Gamma \left(\frac{1}{6}\right) \Gamma \left(\frac{4+a}{6}\right)}{\Gamma \left(\frac{2}{3}\right) \Gamma \left(\frac{1+a}{6}\right)}\right)$ – Mariusz Iwaniuk Dec 27 '23 at 11:34
  • Yes, this closed form had been confirmed by previous answers in the linked post. It involved the digamma function, so this form is clearly expected. –  Dec 27 '23 at 12:02

1 Answers1

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Some clever rearranging and knowledge of special functions and their specific properties completes the proof.

Note that:

$$\mathcal L(\ln(1+\frac{t}{a}))(s) = \frac{e^{sa}}{s}\cdot (-\operatorname{Ei}(-sa))$$

and by definition, we have:

$$-Ei(-x) = \int_{x}^{\infty} \frac{e^{-t}}{t} dt$$

We substitute $u = (6n+1)t$ and $v=(6n+4)t$ and obtain:

$$\mathcal L(I(a))(s) = \sum_{n=0}^{\infty} \frac{1}{s} \cdot e^{(6n+1)s} \cdot \int_{(6n+1)s}^{\infty} \frac{e^{-u}}{u}du - \frac{1}{s} \cdot e^{(6n+4)s} \cdot \int_{(6n+4)s}^{\infty} \frac{e^{-v}}{v}dv $$

Taking inverse Laplace transform on both sides, we have:

$$\ I(a) = \sum_{n=0}^{\infty} \mathcal L^{-1} \big[\frac{e^{(6n+1)s}}{s}\cdot \operatorname{Ei}(-(6n+1)s)\big] - \mathcal L^{-1} \big[\frac{e^{(6n+4)s}}{s}\cdot \operatorname{Ei}(-6n+4)s)\big]$$

$$\ I(a) = \sum_{n=0}^{\infty} \ln(1+\frac{a}{6n+4}) - \ln(1+\frac{a}{6n+1})$$

Or,

$$\ I(a) = \sum_{n=0}^{\infty}\ln \big[\frac{(6n+4)}{(6n+1)} \cdot \frac{(6n+1+a)}{(6n+4+a)}\big]$$

Note that

$$I(0) = \sum_{n=0}^{\infty} \ln(1) = 0$$

Differentiating $I(a)$ w.r.t. $a$, we get:

$$I'(a) = \sum_{n=0}^{\infty} \frac{1}{6n+a+1} - \frac{1}{6n+a+4} $$

$$ = \frac{1}{6} \cdot \sum_{n=0}^{\infty} \frac{1}{n + \frac{a+1}{6}} - \frac{1}{n + \frac{a+4}{6}} $$

$$I'(a) = \frac{1}{6}\cdot \big[\psi(\frac{a+1}{6}) - \psi(\frac{a+4}{6})\big]$$

Integrating from $a=0$ to $a=a$, we get:

$$I(a) = \ln \big(\frac{\Gamma(\frac{a+4}{6}) \cdot \Gamma(\frac{1}{6})}{\Gamma(\frac{a+1}{6}) \cdot \Gamma(\frac{2}{3})}\big)$$

Which is the required result. This follows from the definition of the digamma function, which is the derivative of the logarithm of the Gamma function.

$$ \psi(z) = \frac{d}{dz} (ln(\Gamma(z))$$