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: Write the given integral.
\[ I=\int_0^1 \left(\sum_{k=1}^{\infty}\frac{(\log_e 2)^k x^{k^2-1}}{(k-1)!}\right)dx \]

Step 2: Interchange summation and integration.
Since the series is convergent on \([0,1]\), we write
\[ I=\sum_{k=1}^{\infty}\frac{(\log_e 2)^k}{(k-1)!}\int_0^1 x^{k^2-1}dx \]

Step 3: Evaluate the integral.
\[ \int_0^1 x^{k^2-1}dx=\frac{1}{k^2} \]

Step 4: Substitute this value.
\[ I=\sum_{k=1}^{\infty}\frac{(\log_e 2)^k}{(k-1)!k^2} \]

Step 5: Put \(a=\log_e 2\).
\[ I=\sum_{k=1}^{\infty}\frac{a^k}{(k-1)!k^2} \]
where
\[ a=\log_e 2 \approx 0.6931 \]

Step 6: Compute the numerical value.
\[ I= \frac{a}{1^2} + \frac{a^2}{1!\cdot 2^2} + \frac{a^3}{2!\cdot 3^2} + \frac{a^4}{3!\cdot 4^2} +\cdots \]
\[ I \approx 0.834461 \]

Step 7: Round off to one decimal place.
\[ I \approx 0.8 \] \[ \boxed{0.8} \]