Feynman Differentiation under the Integral Sign

Feynman Integration is a powerful concept that allows for boiling down by normal functions un-integrable formulas. Starting with an un-integrable function as this one

$$\int \_0^1\frac{x-1}{lnx}dx$$

We define this integral as a function of a new simplification-variable $t$ and find a spot for it in the integral, for which, after being differentiated with respect to $t$ the integrand with respect to $x$ may be simplified. Here, we would choose to put it as exponent of $x$ and with the current value of 1, so

$$I(t)=\int \_0^1\frac{x^t-1}{lnx}dx|I(1)$$

When differentiating both sides, the integrand should be simplified if an adequate $t$ was chosen

$$\frac{d}{dt}I(t)=\int \_0^1\frac{\partial }{\partial t}(\frac{x^t-1}{lnx})dx=$$

Differentiating creatively with $x^t=e^{tlnx}$

$$I^{\prime }(t)=\frac{1}{lnx}\frac{\partial }{\partial t}(e^{tlnx}-1)=\frac{x^{t}lnx}{lnx}=x^t$$

The integrand can now be evaluated with respect to $x$

$$I^{\prime }(t)=\int x^{t}dx=\frac{1}{t+1}$$

Another integration, this time with respect to $t$ is required to retrieve the original function $I(t)$ from $I^{\prime }(t)=\frac{d}{dt}I(t)$

$$I(t)=\int \frac{1}{t+1}dt=log|t+1|+C$$

Now notice that we have developed another formula for $I(t)$. This means that we can now set them both equal to each-other and begin finding the constant of integration $C$.

$$I(t)=I(t)\to \int \_0^1\frac{x^t-1}{lnx}dx=log|t+1|+C$$

Now we choose a convenient value for $t$ that will simplify the process. Here, $t=0$ will cause the integrand, and with that the entire left-hand side equation to evaporate to 0. Setting $t=0$

$$0=log|1|+C\to 0=0+C\to C=0$$

We find that $C=0$. We plug this value of $C$ into the, by us developed equation for the integrand, namely $I(t)$ and have solved the un-integrable integral with Feynman integration.

$$I(t)=\int \_0^1\frac{x-1}{lnx}=log|t+1|\to \int \_0^1\frac{x-1}{lnx}=log|2||I(1)$$