How can I approximate $\int_{0}^{n} dx \,\ln(x)$ for large values of $n$?
I am able to get indefinite integration is: $$\int \ln(x) \ dx = x \ \ln(x) - x + C, \quad C\in \Bbb R$$
How can I approximate $\int_{0}^{n} dx \,\ln(x)$ for large values of $n$?
I am able to get indefinite integration is: $$\int \ln(x) \ dx = x \ \ln(x) - x + C, \quad C\in \Bbb R$$
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$$\int^{n}_{0} dx \hspace{0.1cm}\ln(x) = \lim_{a \to 0^{+}} \int^{n}_{a} dx \hspace{0.1cm}\ln(x)\\ \hspace{6cm}= \lim_{a \to 0^{+}} n\hspace{0.1cm} \ln(n) - b\hspace{0.1cm}\ln(a) - n + \ln(a)\\\hspace{6.3cm}= \lim_{a \to 0^{+}}\bigg[ n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + a\hspace{0.1cm}[1 - \ln(a)]\bigg]\\ \hspace{5.8cm}=n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + \lim_{a \to 0^{+}}\frac{1 - \ln(a)}{\frac{1}{a}}\\\hspace{4.2cm}= n\hspace{0.1cm}\ln\bigg(\frac{n}{e}\bigg) + \lim_{a \to 0^{+}}a\\\hspace{2.cm} = n\hspace{0.2cm}\ln\bigg(\frac{n}{e}\bigg)$$