Evaluating $\displaystyle\int_{0}^{\infty} \frac{{x}^{2/3}-{x}^{1/4}}{(1+{x}^{2})\ln(x)} dx$ using the gamma and beta functions

Question

How do I evaluate this integral using the gamma and beta functions? $$\int_{0}^{\infty} \frac{{x}^{2/3}-{x}^{1/4}}{(1+{x}^{2})\ln(x)} dx $$

I don't know what substitution I should make in order to use gamma and beta functions. I tried to make substitution $t=x^2$, but this substitution unfortunately doesn't work.

Answer 1

Define $$I(a) = \int_{0}^{\infty}\frac{x^{a} - x^{\frac{1}{4}}}{(1+x^{2})\ln{x}} \ dx$$ By differentiating under the integral sign, $$I'(a) = \int_{0}^{\infty}\frac{x^{a}}{1+x^{2}} \ dx$$ With a substitution, $$I'(a) = \int_{0}^{\infty}\frac{x^{a}}{1+x^{2}} \ dx = \int_{0}^{\infty}\frac{x^{\frac{a-1}{2}}}{1 + x} \ dx = \textbf{B}\left(\frac{a+1}{2}, 1 - \frac{a+1}{2}\right) = \frac{\pi}{2\cos{\left(\frac{\pi a}{2}\right)}}$$ where $\textbf{B}(z_{1},z_{2})$ is the Beta function. Then, $$I'(a) = \frac{\pi}{2\cos{\left(\frac{\pi a}{2}\right)}} \implies I(a) = \ln\left(\tan{\frac{\pi a}{2}} + \sec{\frac{\pi a}{2}}\right) + C$$ We have that $I(1/4) = 0$, hence, $$C = -\ln\left(\tan{\frac{\pi}{8}} + \sec{\frac{\pi}{8}}\right) = -\ln(-1 + \sqrt{2} + \sqrt{4 - 2\sqrt{2}})$$ Thus, $$I(2/3) = \int_{0}^{\infty}\frac{x^{\frac{2}{3}} - x^{\frac{1}{4}}}{(1+x^{2})\ln{x}} \ dx = \ln\left(\tan{\frac{\pi}{3}} + \sec{\frac{\pi}{3}}\right) -\ln(-1 + \sqrt{2} + \sqrt{4 - 2\sqrt{2}})$$ $$= \ln\frac{2 + \sqrt{3}}{-1 + \sqrt{2} + \sqrt{4 - 2\sqrt{2}}}$$

Answer 2

$$I=\int_{0}^{\infty} \frac{{x}^{2/3}-{x}^{1/4}}{(1+{x}^{2})\ln(x)} dx $$

Let $$J=\int_{0}^{\infty} \frac{{x}^{a}-1}{(1+{x}^{2})\ln(x)} dx,~~~I=J\left(\frac{2}3\right)-J\left(\frac{1}4\right) $$ we have $$\frac{dJ}{da}=\int_{0}^{\infty} \frac{{x}^{a}}{1+{x}^{2}} dx$$

Let $t=x^2$

$$\frac{dJ}{da}=\frac{1}2\int_{0}^{\infty} \frac{t^{\frac{a-1}2}}{1+t} dt=\frac{1}2B\left(\frac{a+1}2,\frac{1-a}2\right)=\frac{\pi}{2\sin\left(\frac{\pi(a+1)}2\right)}$$

we have

$$I=J\left(\frac{2}{3}\right)-J\left(\frac{1}{4}\right)=\int_{1/4}^{2/3} \dfrac{\pi}{2\sin\left(\dfrac{\pi(a+1)}2\right)} da=\ln\left| \dfrac{1+\sin\dfrac{a\pi}2}{\cos\dfrac{a\pi}2}\right|_{1/4}^{2/3} $$

Answer 3

In general, $$ \int_0^{\infty} \frac{x^a-x^b}{\left(1+x^c\right) \ln x}=\ln \left|\frac{\csc \frac{\pi(b+1)}{c}+\cot \frac{\pi(b+1)}{c}}{\csc \frac{\pi(a+1)}{c}+\cot \frac{\pi(a+1)}{c}}\right| $$ Proof:

Feynman’s trick again used here by letting $$I(a)=\int_0^{\infty} \frac{x^a-x ^b}{\left(1+x^c\right) \ln x}$$ where $I(b)=0$.

As usual, differentiating $I(a)$ w.r.t. $a$ yields $$ \begin{aligned} I^{\prime}(a) & =\int_0^{\infty} \frac{x^a}{1+x^c} d x =\frac{\pi}{c} \csc \left(\frac{\pi(a+1)}{c}\right) \end{aligned} $$ where the last step comes from the post.

Integrating $I^{\prime}(a)$ w.r.t $a$ from $b$ to $a$ gives $$ \begin{aligned} I(a)-I(b)&=\frac{\pi}{c} \int_b^a \csc \left(\frac{\pi(t+1)}{c}\right) d t \\ I(a)&=-\left[\ln \left|\csc\left(\frac{\pi(t+1)}{c}\right)+\cot \left(\frac{\pi(t+1)}{c}\right)\right| \right]_b^a\\&= \ln \left|\frac{\csc \frac{\pi(b+1)}{c}+\cot \frac{\pi(b+1)}{c}}{\csc \frac{\pi(a+1)}{c}+\cot \frac{\pi(a+1)}{c}}\right|\\\\ \end{aligned} $$ For examples,

$$\int_0^{\infty} \frac{x^{\frac23}-x^{\frac14}}{\left(1+x^2\right) \ln x}= \ln \left(\frac{\sec \frac{\pi}{3}+\tan \frac{\pi}{3}}{\sec \frac{\pi}{8}+\tan \frac{\pi}{8}}\right) = \ln \left(\frac{2+\sqrt{3}}{\sqrt{2(2-\sqrt{2})}+\sqrt{2}-1}\right) $$

$$\int_0^{\infty} \frac{\sqrt x-\sqrt[3]x}{\left(1+x^3\right) \ln x}= \ln \left(\operatorname{csc} \frac{4 \pi}{9}+\cot \frac{4 \pi}{9}\right) \approx 0.1754258$$