Science:Math Exam Resources/Courses/MATH101/April 2013/Question 12 (a)/Solution 1

Since it is difficult to evaluate this integral directly, we try using the comparison test instead to prove divergence. Notice that for large $\displaystyle x$, the integrand behaves like $\displaystyle 1/x$. Since $\int _{2}^{\infty }1/x\ dx$ diverges, we will try to show that our integrand is larger than a constant times $\displaystyle 1/x$.

First, since $\displaystyle \sin x\geq -1$, we have

$\displaystyle {\frac {x+\sin x}{1+x^{2}}}\geq {\frac {x-1}{1+x^{2}}}\,.$

Now for $\displaystyle x\geq 2$, we have

$x-1\geq {\frac {x}{2}}\quad$

and

$1+x^{2}\leq 2x^{2}.$

(In fact, the latter is even true for $x\geq 1$.)

Hence, over the interval of integration, we have the lower bound

${\frac {x+\sin x}{1+x^{2}}}\geq {\frac {x}{2}}\cdot {\frac {1}{2x^{2}}}={\frac {1}{4x}}\,.$

Finally, since

$\displaystyle \int _{2}^{\infty }{\frac {1}{4x}}\,dx={\frac {1}{4}}\int _{2}^{\infty }{\frac {1}{x}}\,dx$

diverges, we conclude that the integral in question diverges as well by the comparison test.