# Show that $\sum_{n=1}^{\infty}\frac{(-1)^n}{n}\int_{1}^{\infty}\frac{(\sqrt[t]{t}-1)^{n}t}{2^{t}}dt=\frac{\operatorname{Ei}(-\ln2)}{\ln2}$

Let $$T(x)=\int_{1}^{\infty}\frac{(\sqrt[t]{t}-1)^{x}t}{2^{t}}dt$$ Then show that;

$$\sum_{n=1}^{\infty}\frac{(-1)^nT(n)}{n}=\frac{\operatorname{Ei}\left(-\ln2\right)}{\ln2}$$

Where $$\operatorname{Ei}(x )=-\int_{-x}^{\infty}\frac{e^{-t}}{t}dt$$, the Exponential Integral.

• Your own attempts?.. (Did you try the obvious $\sum\int\mapsto\int\sum$?) Jan 17 '20 at 12:17

\begin{align} \sum_{n=1}^\infty\frac{(-1)^nT(n)}{n} &=\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_1^\infty\frac{(\sqrt[t]{t}-1)^nt}{2^t}\mathrm{d}t\\ &=\int_1^\infty\frac{t}{2^t}\sum_{n=1}^\infty\frac{(1-\sqrt[t]{t})^n}{n}\mathrm{d}t\\ &=\int_1^\infty-\frac{t}{2^t}\ln{(\sqrt[t]{t})}\mathrm{d}t\\ &=-\int_1^\infty\frac{\ln{(t)}}{2^t}\mathrm{d}t\\ &=-\frac1{\ln{(2)}}\int_{\ln{(2)}}^\infty\frac{\ln{(u/\ln{(2)})}}{e^u}\mathrm{d}u\\ &=-\frac1{\ln{(2)}}\int_{\ln{(2)}}^\infty(e^{-u}\ln{(u)}-\ln{(\ln{(2)})}e^{-u})\mathrm{d}u\\ &=-\frac1{\ln{(2)}}\left[\text{Ei}(-u)-e^{-u}\ln{(u)}+\ln{(\ln{(2)})}e^{-u}\right]_{\ln{(2)}}^\infty\\ &=-\frac1{\ln{(2)}}\left[(0-0+0)-\left(\text{Ei}(-\ln{(2)})-\frac{\ln{(\ln{(2)})}}{2}+\frac{\ln{(\ln{(2)})}}{2}\right)\right]\\ &=\frac{\text{Ei}(-\ln{(2)})}{\ln{(2)}}\\ \end{align}