Question

\( \displaystyle \int_0^1 \left(\sum_{k=1}^{\infty}\frac{(\log_e 2)^k x^{k^2-1}}{(k-1)!}\right)dx = \underline{} \) rounded off to one decimal place.

Hint:

For integrals containing infinite series, first integrate \(x^{k^2-1}\) term-wise, then evaluate the resulting numerical series.

Answer 1

Correct Answer: 0.8

Step 1: Swap the sum and integral.
The series converges nicely on $[0,1]$, so we may integrate term by term.
\[ I=\sum_{k=1}^{\infty}\frac{(\log 2)^k}{(k-1)!}\int_0^1 x^{k^2-1}\,dx \]

Step 2: Do the inner integral.
\[ \int_0^1 x^{k^2-1}\,dx=\frac{1}{k^2} \]

Step 3: Put it back.
\[ I=\sum_{k=1}^{\infty}\frac{(\log 2)^k}{(k-1)!\,k^2} \]

Step 4: Let $a=\log 2$.
Here $a=\log 2\approx 0.6931$, so the terms shrink fast because of the $(k-1)!$ in the bottom.

Step 5: Add a few terms.
\[ I\approx a+\frac{a^2}{4}+\frac{a^3}{18}+\frac{a^4}{96}+\cdots\approx 0.8345 \]

Step 6: Round it.
To one decimal place the value is $0.8$.
\[ \boxed{0.8} \]