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