Using Cauchy product for an integral

Question

To evaluate $$ \int_0^1 e^x \ln(x+1)dx $$ I was thinking about using the cauchy product of the taylor series of $e^x$ and $\ln(x+1)$. We know that $$ e^x = \sum_{n=0}^\infty \frac{x^n}{n!} $$ and $$ \ln(x+1) = \sum_{m=0}^\infty \frac{(-1)^{m+1}x^m}{m} $$ So using: $$ \left(\sum_{n=0}^\infty a_n x^n\right) \cdot \left(\sum_{m=0}^\infty b_m x^m \right) = \sum_{n=0}^\infty \sum_{m=0}^n (a_m b_{n-m}) x^n $$ We have: $$ \int_0^1 e^x \ln(x+1)dx = \int_0^1 \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{m\cdot m!} b_{n-m}\right) x^n dx $$ I switch the integration and summation bounds but I do not know how to justify this: $$ \int_0^1 e^x \ln(x+1)dx = \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{(n-m)\cdot m!}\right) \int_0^1 x^n dx $$ Obtaining: $$ \int_0^1 e^x \ln(x+1)dx = \sum_{n=0}^\infty \sum_{m=0}^n \left(\frac{(-1)^{1 + n - m}}{(n-m)(n+1)\cdot m!} \right) $$ I tried to plug this into desmos but it does not work.

Answer 1

Your integral is

$$ \int_0^1 e^x \ln(x+1) dx=e\ln(2)-\int_0^1 \dfrac{e^x}{x+1}dx=e\ln(2)-\int_1^2e^{-1}\dfrac{e^u}{u}du$$

Which is $$ e\ln(2)-e^1 (Ei(1)-Ei(2))$$

Where $Ei$ is the exponential integral :

https://mathworld.wolfram.com/En-Function.html