Double integration problem

I am going to attempt a solution which in no way is a proof. I would like to assume the convergence of the integral as I am unable to prove it.

First I want to substitute $xy=t$ so $xdy = dt$ .

So we have $\displaystyle \int_{0}^{1}\int_{0}^{y}\frac{\frac{t}{y}-1}{(1-t)\ln(t)}\frac{1}{y}dtdy$

Now I want to change the order of integration:-

So we have:-

$\displaystyle \int_{0}^{1}\int_{t}^{1}\frac{\frac{t}{y}-1}{(1-t)\ln(t)}\frac{1}{y}dydt$

$\displaystyle = \int_{0}^{1}\frac{1-t+\ln(t)}{(1-t)\ln(t)}dt$

Here comes the fishy steps.

$\displaystyle = \int_{0}^{1}\frac{1}{1-t}dt +\int_{0}^{1}\frac{1}{\ln(t)} dt$ [Note that this step is invalid if the integral was not convergent]

$\displaystyle = \int_{0}^{1}\frac{1}{t} dt +\int_{0}^{1}\frac{1}{\ln(t)}dt$ [Note that this step is also invalid if the integral was not convergent]

Now let us rewrite this as :-

$\large = \lim_{\varepsilon\to 0}\int_{\varepsilon}^{1}\frac{1}{t} dt +\int_{0}^{1}\frac{1}{\ln(t)}dt$

$\large = \lim_{\varepsilon\to 0}-\ln(\varepsilon) +\int_{0}^{1}\frac{1}{\ln(t)}dt$

Now let us use the substitution $t=e^{-z}$ for the log integral and then write it in limit form:-

$\large = \lim_{\varepsilon\to 0}\left(-\ln(\varepsilon)\right) -\lim_{\varepsilon\to 0\,X\to\infty}\left(\int_{\varepsilon}^{X}\frac{e^{-z}}{z}dz\right)$

Now in the integral I want to apply by parts with $\frac{1}{z}$ as the integrable function. So we have:-

$\large = (\lim_{\varepsilon\to 0}-\ln(\varepsilon)) +\lim_{\varepsilon\to 0\,X\to\infty}\left(\ln(\varepsilon)e^{-\varepsilon}-e^{-X}\ln(X)-\int_{\varepsilon}^{X}e^{-z}(\ln(z))dz\right)$

Now $\large \lim_{\varepsilon\to 0\,X\to\infty}\int_{\varepsilon}^{X}e^{-z}(\ln(z))dz$ is a well-known integral and it is $\psi(1)$ . Where $\psi$ denotes the digamma function and it converges to $-\gamma$ . Where $\gamma$ is the Euler-Mascheroni constant .

And we have $\displaystyle \lim_{\varepsilon\to 0\,X\to\infty} e^{-X}\ln(X)$ tend to $0$ .

So what we have left is $\displaystyle \lim_{\varepsilon\to 0} (1-e^{-\varepsilon})(\ln(\varepsilon)) + \gamma$

Rewriting the limit as $\displaystyle \lim_{\varepsilon\to 0} \frac{(1-e^{-\varepsilon})}{\varepsilon} \cdot \varepsilon(\ln(\varepsilon)) + \gamma$

Now $\displaystyle \lim_{\varepsilon\to 0} \frac{(1-e^{-\varepsilon})}{\varepsilon}$ tends to $1$ and $\displaystyle \lim_{\varepsilon\to 0}\varepsilon^{n}\ln(\varepsilon)$ $\to 0$ for all natural numbers $n$ .

So we have our answer as $\gamma = 0.5772$ .

I understand that this is in no way a proof. 90% of the steps has no meaning until convergence is proven . But still it was my attempt and I wanted to share it since there was no solution posted . I hope it would be interesting for the one's viewing it .

Just apply the hadjicostas formula. And by assigning the values we will get the famous euler mascheroni constant. Which is some what 0.5772.....