# Definite integral with a strange power

I've been struggling for a while now on evaluating this disgusting integral: $$(\ln2)\int_0^{(\ln2)^{1/\ln2}}2^{\ln x}\cdot x^\left(\frac{x^{\ln2}+1}{\ln x}-1\right)dx$$ My maths teacher gave our class this question a while ago, and he said that we should be able to do it (I am in high-school, and we have only been taught a fairly basic level of integration). So today I spent many hours applying every integration technique I know to this monster, but I got absolutely nowhere. It got to a point where I couldn't think of another variable to use as a substitution because I had already made so many.

I eventually decided to plug this into an integral calculator and received a surprisingly nice result of $$e$$, however there was no further information and so I was not able to view any of the steps in how they got there.

I am so stuck on this problem :(

Does anyone know how they got there? What are the steps in finding its indefinite form?

• This seems like a mean question but in any case the first substitution I'd try is $u = \ln x$, especially given that $-1$ in the exponent. Remember that $x^{\ln 2} = e^{\ln 2 \ln x}$. Good luck! Oct 17, 2022 at 8:34
• @QiaochuYuan I tried this substitution, but when I subbed in the relevant u values for x, I ended up with $\ln{0}=u$ which is undefined, right? Oct 17, 2022 at 8:43
• @SickNutmeg It must be seen as an improper integral to $-\infty$ Oct 17, 2022 at 8:46
• We have 4 or 5 more or less identical answers! Presumably all typed at the same time Oct 17, 2022 at 9:29
• If $w = x^{\log_e 2},~\text {then} \log_e2= \dfrac{\log_u w}{\log_u x}\text{for any arbitrary real u.}$ Oct 17, 2022 at 9:29

we start by a well known notion $$a=e^{\ln{(a)}}$$ then $$\frac{x^{\ln{(2)}}+1}{\ln x}-1$$ $$=\frac{x^{\ln{(2)}}+1-\ln{(x)}}{\ln{(x)}}$$ Now $$x^{\frac{x^{\ln{(2)}}+1}{\ln x}-1}$$ $$=\left(e^{\ln{(x)}}\right)^\frac{x^{\ln{(2)}}+1-\ln{(x)}}{\ln{(x)}}$$ $$=e^{[x^{\ln{(2)}}+1-\ln{(x)}]}$$ $$=\frac{e^{[x^{\ln{(2)}}+1]}}{e^{\ln{(x)}}}$$ $$=\frac{e^{[x^{\ln{(2)}}+1]}}{x}$$

and since $$2^{\ln{(x)}}=\left(e^{\ln{(2)}}\right)^{\ln{(x)}}=\left(e^{\ln{(x)}}\right)^{\ln{(2)}}=x^{\ln{(2)}}$$ we get

$$2^{\ln{(x)}}\frac{x^{\ln{(2)}}+1}{\ln x}-1$$ $$=x^{\ln{(2)}}\frac{e^{[x^{\ln{(2)}}+1]}}{x}$$ $$=x^{[\ln{(2)}-1]}e^{[x^{\ln{(2)}}+1]}$$

From here on taking

$$u=x^{\ln{(2)}}+1$$ $$\Rightarrow du=\ln{(2)}x^{[\ln{(2)}-1]}dx$$

should change your integral into an $$e^u$$ integral.

• Thank you - This is the only one I truly understood, but also thanks to everyone else who gave an answer, it is greatly appreciated. Oct 17, 2022 at 9:49
• You are more than welcome 😊
– Sam
Oct 17, 2022 at 9:57

This falls to some well-motivated $$u$$-substitutions: \begin{align*} \ln2\int 2^{\ln x} x^{\left(\frac{x^{\ln2}+1}{\ln x}-1\right)} dx &= \ln 2 \int 2^t e^{2^t+1} dt & \textrm{let x=e^t} \\ &= \int e^{s+1} ds & \textrm{let t=\log_2 s}. \end{align*} In total this amounts to the $$u$$-substitution $$x = s^{\log_2 e}$$, which is bit more difficult to see.

Denote $$f(x)=x^{\frac{x^{\ln 2}+1}{\ln x}}, x>0.$$ Let's calculate the derivative of $$f$$. Using the standard method, we differentiate the both sides of $$\ln f(x)=\frac{x^{\ln 2}+1}{\ln x}\ln x=x^{\ln 2}+1,\tag{*}$$ to get $$\frac{f'(x)}{f(x)}=\ln 2\cdot x^{\ln 2-1},$$ so $$f'(x)=\ln 2\cdot x^{\ln 2-1}f(x).$$ We need to compute the integral $$I=\int_0^{(\ln2)^{1/\ln2}}\ln 2\cdot2^{\ln x}x^{-1}f(x)\,dx.$$ Indeed, we have $$2^{\ln x}=e^{\ln2\cdot \ln x}= x^{\ln 2}$$, so the integral $$I=\int_0^{(\ln2)^{1/\ln2}}\ln 2\cdot x^{\ln 2-1}f(x)\,dx=\int_0^{(\ln2)^{1/\ln2}}f'(x)\,dx=f\left((\ln2)^{1/\ln2}\right)-f(0+).$$ Now, we can easily get $$f\left((\ln2)^{1/\ln2}\right)=2e$$ and $$f(0+)=e$$ from $$(*)$$.

$$\newcommand{\d}{\mathrm{d}}$$This is an extreme example of having more bark than bite. This is also a highly artificial integral, and everything just ‘magically’ works out.

You must convert everything to exp-log form. Begin: \begin{align}I&:=(\ln2)\int_0^{(\ln2)^{1/\ln2}}2^{\ln x}x^{\left(\frac{x^{\ln2}+1}{\ln x}-1\right)}\,\d x\\&=(\ln2)\int_0^{\exp(\ln\ln2/\ln2)}\exp\left((\ln x)(\ln2)+e^{(\ln2)(\ln x)}+1-\ln x\right)\,\d x\\&=\int_{-\infty}^{\ln\ln2}\exp(t+e^t+1)\,\d t\end{align}

Note $$x^{a/\ln x}=\exp((\ln x)(a/\ln x))=\exp(a)$$. I also substituted $$t=(\ln2)(\ln x)$$, you can check the details. That’s a natural substitution to make here.

Finally, let $$\omega=e^t$$, so that $$\d\omega=e^t\,\d t$$ and the RHS is present in our integrand. So we get: $$\int_0^{\ln2}\exp(\omega+1)\,\d\omega$$Which is just $$e$$.

Rewrite the integrand as $$\exp(f(\ln x))$$ and simplify: $$(\ln2)\int_0^{(\ln2)^{1/\ln2}}\exp(e^{\ln 2\ln x}+1-\ln x+\ln2\ln x)\,dx$$ $$=e\int_0^{(\ln2)^{1/\ln2}}\exp(e^{\ln 2\ln x})e^{\ln2\ln x}(\ln2)\frac1x\,dx$$ We see the factors produced by applying the chain rule when differentiating $$\exp(e^{\ln2\ln x})$$. The integrand has an elementary derivative: $$=e[\exp(e^{\ln2\ln x})]_0^{(\ln2)^{1/\ln2}}=e(2-1)=e$$