Mathematics/Calculus/Corner cases

Derivatives

The derivative of ${\frac {d}{dx}}x^{x}$

Question

What is ${\frac {d}{dx}}x^{x}$?

Solution 1

• Let $y=x^{x}$.
• Take $\ln$ of both sides: $\ln y=x\ln x$.
• Differentiate both sides: ${\frac {d}{dx}}\ln y={\frac {d}{dx}}x\ln x$.
• Apply the chain rule on the left-hand side: ${\frac {d}{dx}}\ln y={\frac {1}{y}}\cdot {\frac {dy}{dx}}$.
• Apply the product rule on the right-hand side: ${\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1$.
• Putting it together, we have ${\frac {1}{y}}\cdot {\frac {dy}{dx}}=\ln x+1$.
• Hence ${\frac {dy}{dx}}=y(\ln x+1)=x^{x}(\ln x+1)$.

Solution 2

• Note that $x=e^{\ln x}$, so $x^{x}=(e^{\ln x})^{x}=e^{x\ln x}$.
• Applying the chain rule, ${\frac {d}{dx}}x^{x}={\frac {d}{dx}}e^{x\ln x}=e^{x\ln x}{\frac {d}{dx}}x\ln x$.
• Applying the product rule, ${\frac {d}{dx}}x\ln x=1\cdot \ln x+x\cdot {\frac {1}{x}}=\ln x+1$.
• Therefore ${\frac {d}{dx}}x^{x}=e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)$.

Integrals

The integral $\int x^{x}\,dx$

Question

What is $\int x^{x}\,dx$?

Solution

• We can write $x^{x}$ as $(e^{\ln x})^{x}=e^{x\ln x}$.
• Consider the series expansion of $e^{x\ln x}$:

   $e^{x\ln x}=1+(x\ln x)+{\frac {(x\ln x)^{2}}{2!}}+{\frac {(x\ln x)^{3}}{3!}}+\ldots +{\frac {(x\ln x)^{i}}{i!}}+\ldots =\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}$.

• We can interchange the integration and summation (we can recognize this as a special case of the Fubini/Tonelli theorems) and write

   $\int x^{x}\,dx=\int \left(\sum _{i=0}^{\infty }{\frac {(x\ln x)^{i}}{i!}}\right)\,dx=\sum _{i=0}^{\infty }\left(\int {\frac {(x\ln x)^{i}}{i!}}\,dx\right)=\sum _{i=0}^{\infty }\left({\frac {1}{i!}}\int x^{i}(\ln x)^{i}\,dx\right).$

Limits

The limit of $\lim _{x\rightarrow 0^{+}}x^{x}$

Question

What is $\lim _{x\rightarrow 0^{+}}x^{x}$?

Solution

• Note that $x=e^{\ln x}\$,so\$x^{x}=(e^{\ln x})^{x}=e^{x\ln x}$.
• We can further rewrite this as $x^{x}=e^{x\ln x}=e^{\frac {\ln x}{\frac {1}{x}}}$.
• As long as $f$ is continuous and the limit of $g$ exists at the point in question, the limit will commute with composition:

$\lim _{x\rightarrow t}f(g(x))=f(\lim _{x\rightarrow t}g(x)).$ In our case, $e(\cdot )$ is continuous, so $\lim _{x\rightarrow 0^{+}}x^{x}=e^{\lim _{x\rightarrow 0^{+}}{\frac {\ln x}{\frac {1}{x}}}}.$

• The question, then, is what is $\lim _{x\rightarrow 0^{+}}{\frac {\ln x}{\frac {1}{x}}}$.
• As $x\rightarrow 0^{+}$, $\ln x\rightarrow -\infty \$,\${\frac {1}{x}}\rightarrow +\infty$. In this situation we can apply l'Hôpital's rule:

$\lim _{x\rightarrow 0^{+}}{\frac {\ln x}{\frac {1}{x}}}=\lim _{x\rightarrow 0^{+}}{\frac {{\frac {d}{dx}}\ln x}{{\frac {d}{dx}}{\frac {1}{x}}}}={\frac {\frac {1}{x}}{-{\frac {1}{x^{2}}}}}={\frac {{\frac {1}{x}}\cdot x^{2}}{-{\frac {1}{x^{2}}}\cdot x^{2}}}=-x.$

• Hence $\lim _{x\rightarrow 0^{+}}x^{x}=e^{0}=1$.

The limits of $\lim _{x\rightarrow +\infty }x\sin {\frac {1}{x}}$ and $\lim _{x\rightarrow -\infty }x\sin {\frac {1}{x}}$

Question

What are Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x}} and Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x}} ?

Solution

• Let us rewrite $x\sin {\frac {1}{x}}\$as\${\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}$.
• As Failed to parse (unknown function "\tendsto"): {\displaystyle x \tendsto +\infty$, $\frac{1}{x} \tendsto 0} and Failed to parse (unknown function "\tendsto"): {\displaystyle x \sin \frac{1}{x} \tendsto 0} .
• We have ``${\frac {0}{0}}$, so we can apply l'Hôpital's rule.
• Differentiating the numerator in ${\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}$, we obtain $\left(\cos {\frac {1}{x}}\right)\left(-{\frac {1}{x^{2}}}\right)$.
• Differentiating the denominator in ${\frac {\sin {\frac {1}{x}}}{\frac {1}{x}}}$, we obtain $-{\frac {1}{x^{2}}}$.
• Thus

Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto +\infty} x \sin \frac{1}{x} = \lim_{x \tendsto +\infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}} = \lim_{x \tendsto +\infty} \frac{\left(\cos \frac{1}{x}\right) \left(-\frac{1}{x^2}\right)}{-\frac{1}{x^2}} = \lim_{x \tendsto +\infty} \cos \frac{1}{x} = 1. }

• Similarly we can find that Failed to parse (unknown function "\tendsto"): {\displaystyle \lim_{x \tendsto -\infty} x \sin \frac{1}{x} = 1} .