Indian Statistical Institute B.Math & B.Stat : Integration

Let $n$ be positive integers and $[.]$ be the floor function. Evaluate the integral \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx\] Since \( 0<1<2<3< \dots < n-1 < n \) we have \( 0^{\frac{1}{k}}<1^{\frac{1}{k}}<2^{\frac{1}{k}}<3^{\frac{1}{k}}< \dots < (n-1)^{\frac{1}{k}} < n^{\frac{1}{k}} \) $$$$ Therefore \[ \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx \] Now for a fixed $i$ in the range, consider the integral \[\int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx \] \[ i^{\frac{1}{k}} < x < (i+1)^{\frac{1}{k}} \implies i < x^k < i+1 \implies x^k = i + \epsilon \quad where \quad 0 < \epsilon < 1\] \[ therefore \quad [x^k +n] = [i + \epsilon+n]= [i+n+ \epsilon] = i+n \] \[ therefore \quad \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} [x^k + n] dx = \int_{i^{\frac{1}{k}}}^{(i+1)^{\frac{1}{k}}} (i+n) dx = (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ Thus \quad \int_{0}^{n^{\frac{1}{k}}} [ x^k + n] dx = \sum_{i=0}^{n-1} \int_{i^{\frac{1}{k}}}^{(i+1){\frac{1}{k}}} [x^k + n] dx = \sum_{i=0}^{n-1} (i+n)\bigg((i+1)^{\frac{1}{k}}-i^{\frac{1}{k}}\bigg) \] \[ = (n+0)\bigg(1^{\frac{1}{k}}-0^{\frac{1}{k}}\bigg) + (n+1)\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+ \dots +(n+n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg) + (n+n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \] \( = n \bigg( 1^{\frac{1}{k}}-0^{\frac{1}{k}} +2^{\frac{1}{k}}-1^{\frac{1}{k}}+\dots+(n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}+n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}} \bigg) \) \(+ 1\bigg(2^{\frac{1}{k}}-1^{\frac{1}{k}}\bigg)+2\bigg(3^{\frac{1}{k}}-2^{\frac{1}{k}}\bigg)\dots+(n-2)\bigg((n-1)^{\frac{1}{k}}-(n-2)^{\frac{1}{k}}\bigg)+(n-1)\bigg(n^{\frac{1}{k}}-(n-1)^{\frac{1}{k}}\bigg) \) $$$$ \( = n \times n^{\frac{1}{k}}+2^{\frac{1}{k}}-1^{\frac{1}{k}}+2 \times 3^{\frac{1}{k}}-2 \times 2^{\frac{1}{k}}+ \dots +(n-2)(n-1)^{\frac{1}{k}}-(n-2)(n-2)^{\frac{1}{k}}+ (n-1)n^{\frac{1}{k}}-(n-1)(n-1)^{\frac{1}{k}} \) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}+(n-1)n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n \times n^{\frac{1}{k}}\) $$$$ \( =n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}+n^{\frac{1+k}{k}}\) $$$$ \( =2n^{\frac{1+k}{k}}-1^{\frac{1}{k}}-2^{\frac{1}{k}}-\dots-(n-1)^{\frac{1}{k}}-n^{\frac{1}{k}}\) $$$$ \( = 2n^{\frac{1+k}{k}} - \sum_{i=1}^{n} i^\frac{1}{k} \)