Question

If $1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2} - .........\infty = 0.9159$ and if $I = \int_{0}^{\infty} \frac{x}{e^x+e^{-x}}dx$ then the value of $[10I]$ is equal to (where [.] denotes G.I.F)

Answer 1

9

Answer 2

To evaluate the integral $I = \int_{0}^{\infty} \frac{x}{e^x+e^{-x}}dx$, we first simplify the integrand.

The denominator $e^x+e^{-x}$ can be written as $2 \cosh x$. So, $I = \int_{0}^{\infty} \frac{x}{2 \cosh x} dx = \frac{1}{2} \int_{0}^{\infty} \frac{x}{\cosh x} dx$.

Next, we expand $\frac{1}{\cosh x}$ into a series. $\frac{1}{\cosh x} = \frac{2}{e^x+e^{-x}} = \frac{2e^{-x}}{1+e^{-2x}}$. Using the geometric series expansion $\frac{1}{1+r} = 1-r+r^2-r^3+...$ for $|r|<1$, with $r=e^{-2x}$: $\frac{1}{1+e^{-2x}} = 1 - e^{-2x} + e^{-4x} - e^{-6x} + ...$ This expansion is valid for $x>0$, as $e^{-2x} < 1$. Substituting this back: $\frac{1}{\cosh x} = 2e^{-x} (1 - e^{-2x} + e^{-4x} - e^{-6x} + ...)$ $\frac{1}{\cosh x} = 2 (e^{-x} - e^{-3x} + e^{-5x} - e^{-7x} + ...)$ $\frac{1}{\cosh x} = 2 \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)x}$.

Now, substitute this series into the integral for $I$: $I = \frac{1}{2} \int_{0}^{\infty} x \left( 2 \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)x} \right) dx$ $I = \int_{0}^{\infty} x \left( \sum_{n=0}^{\infty} (-1)^n e^{-(2n+1)x} \right) dx$ We can interchange the integral and the summation (due to uniform convergence of the series for $x \ge \epsilon > 0$): $I = \sum_{n=0}^{\infty} (-1)^n \int_{0}^{\infty} x e^{-(2n+1)x} dx$.

Now we need to evaluate the integral $\int_{0}^{\infty} x e^{-ax} dx$. This integral can be solved using integration by parts or by using the Laplace transform. Using Laplace transform definition $\mathcal{L}\{f(t)\}(s) = \int_0^\infty e^{-st} f(t) dt$. For $f(t)=t$, $\mathcal{L}\{t\}(s) = \frac{1}{s^2}$. So, $\int_{0}^{\infty} x e^{-ax} dx = \frac{1}{a^2}$.

Substitute $a = (2n+1)$ into the result of the integral: $\int_{0}^{\infty} x e^{-(2n+1)x} dx = \frac{1}{(2n+1)^2}$.

Now substitute this back into the sum for $I$: $I = \sum_{n=0}^{\infty} (-1)^n \frac{1}{(2n+1)^2}$ Expanding the sum: $I = (-1)^0 \frac{1}{(2(0)+1)^2} + (-1)^1 \frac{1}{(2(1)+1)^2} + (-1)^2 \frac{1}{(2(2)+1)^2} + ...$ $I = \frac{1}{1^2} - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2} + \frac{1}{9^2} - ...$

This is exactly the series given in the problem statement: $1-\frac{1}{3^2}+\frac{1}{5^2}-\frac{1}{7^2}+\frac{1}{9^2} - .........\infty = 0.9159$. Therefore, $I = 0.9159$.

The question asks for the value of $[10I]$, where $[.]$ denotes the Greatest Integer Function (G.I.F). $10I = 10 \times 0.9159 = 9.159$. $[10I] = [9.159]$. The greatest integer less than or equal to 9.159 is 9.

The final answer is 9.