Integral in an integral in an integral in an ... forever

Calculus Level 4

$...\int_1^x (((((\ln x+1) \int_1^x ( (((\ln x + 1) \int_1^x ( ((\ln x + 1) \int_1^x ((\ln x + 1) \int_1^x ( \ln x + 1) dx ) dx ) dx ) dx ) dx ....$

A series is constructed above such that

the first term is the innermost integral $S_1 = \int_1^x (\ln x+1) dx;$
the second term is the two innermost integrals $S_2 = \int_1^x \big((\ln x+1)\int_1^x (\ln x+1) dx \big)\, dx;$
the third term is the three innermost integrals $S_3 = \int_1^x \Big((\ln x+1)\int_1^x \big((\ln x+1)\int_1^x (\ln x+1) dx \big)\, dx \Big)\, dx,$ etc.

$S_n$ for $n > 0$ can be written in the form $\frac 1{A!} x^B \ln^C(x)$ , where $A$ , $B$ , and $C$ are integers.

Find the value of $A^2BC-101$ for $S_{7}.$

1 solution

Timothy Cao
Apr 4, 2018

Begin with evaluating the first few values of the set, ignoring the constants of integration.

$s_{1} = xlnx$

$s_{2} = \frac{1}{2}x^{2}ln^{2}x$

$s_{3} = \frac{1}{6}x^{3}ln^{3}x$

$s_{4} = \frac{1}{24}x^{4}ln^{4}x$

$s_{5} = \frac{1}{120}x^{5}ln^{5}x$

Here we can notice a pattern which we can generalize:

$s_{n} = \frac{1}{n!}x^{n}ln^{n}x$

Now we define the values for $A, B, C$ :

$A = B = C = n$

Now, plugging this into $A^2BC-101$

$7^{4}-101 = 2300$

Special note: Thank you for Jon Haussmann for correcting an error! The problem should be fixed now.