In the integration formula $\int dx/x = log x + c$, Is the log natural or log base 10? The formula appears in many problems and i just got a problem wrong for apparently using the wrong log. Could you please enlighten me about the right log to be used in integration.
In the integration formula $\int dx/x = log x + c$, Is the log natural or log base 10
logarithms
Best Answer
Consider the differential equation $y'=y$ with $y(0)=1$ on $\mathbb{R}$. It can be proven that this has a unique solution. We define the exponential function on $\mathbb{R}$ to be equal to this solution. We denote it as $\exp$, but many people abuse notation and denote it $e^x$, which in this context, is fine, as long as everyone knows what is being meant. Now, it can be proven that $\exp$ is function whose range is $\mathbb{R}^+$. In other words, $\exp[\mathbb{R}]=\mathbb{R}^+$. It can also be proven that $\exp$ is a function that has a compositional inverse. This compositional inverse has a name: the natural logarithm, and we denote it $\ln$, or $\log$. Of course, one must remember that $\ln[\mathbb{R}^+]=\mathbb{R}$. The relationship that this function has with other logarithmic functions is complicated, but in some sense, it is the logarithmic function, due to the fact that it can be defined independently of any notion of powers. Hence the name "natural" logarithm. In terms of other logarithmic functions, the natural logarithm can be interpreted as being base $e$, where $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$.
Now, given the equation $\exp'=\exp$, it can be proven via the chain rule that $\ln'(x)=\frac{1}{x}$ for every $x\in\mathbb{R}^+$. This is because $(\exp\circ\ln)'(x)=1$, yet by the chain rule, $(\exp\circ\ln)'(x)=(\exp'\circ\ln)(x)\ln'(x)=(\exp\circ\ln)(x)\ln'(x)=x\ln'(x)$, so $x\ln'(x)=1$, implying $\ln'(x)=\frac{1}{x}$. Because of this, $\ln(x)$ is an antiderivative of $\frac{1}{x}$ on $\mathbb{R}^+$.
However, this is not the complete story. Consider the function $\ln^-$ defined by $(\ln^-)(x)=\ln(-x)$ for every $\mathbb{R}^-$. It turns out that $(\ln^-)'(x)=\frac{1}{x}$ is true on $\mathbb{R}^-$ as well! So to talk about the antiderivatives of $\frac{1}{x}$ properly, we need to consider functions $f_{A,B}:\mathbb{R}\setminus\{0\}\rightarrow\mathbb{R}$ such that $$f_{A,B}(x)=\begin{cases} \ln(-x)+A & x\lt0 \\ \ln(x)+B & x\gt0 \end{cases}.$$ These functions satisfy the property that $f_{A,B}'(x)=\frac{1}{x}$ for every $x\in\mathbb{R}\setminus\{0\}$. And in fact, this is every function that satisfies the equation on that domain, there are no other functions. So you can say $$\int\,\frac{\mathrm{d}x}{x}=f_{A,B}(x).$$ If you want to limit yourself to $x\gt0$, though, then $$\int\,\frac{\mathrm{d}x}{x}=\ln(x)+B.$$ Hopefully, this covers everything.