1. Hard Integration Q

Consider the definite integral $D=\int_{-\pi}^{\pi}\cos\lambda x\cos nx\,dx$ where n is a positive integer and lamda is a positive real number.

Given also that $D=\left\{\begin{matrix}
\pi & for\,\lambda =n\\
0 & for\,\lambda \, a positive integer\,,\lambda \neq n\\
\frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2} & for\,\lambda \,not an integer
\end{matrix}\right.$

Show that when $0< \lambda < n,\,|D|< \pi$ and when $\lambda > n+\frac{1}{2},\,|D|< 3$

Help of any kind would be greatly appreciated.

2. Would the first part of the question have something to do with limits? We can see that $0<|\sin(\lambda \pi )|<1$ and so the maximum value for the expression is obviously the limit as lamda approaches n since $\frac{2\lambda}{\lambda ^2-n^2}$ approaches infinite. However, im not sure how to evaluate the limit of the entire expression $\frac{(-1)^n2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}$.

3. I'll attempt the second half.
Let $\lambda > n+0.5$

then

$\lambda^2-n^2 > (n+.5)^2-n^2=n+.25$

therefore,

$|\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<
|\frac{2(n+.5) \sin(\lambda \pi )}{n+.25}|\leq
|\frac{2(n+.5) }{n+.25}|$

Now
$
\frac{(n+.5) }{n+.25}$
is a strictly decreasing function and n is an integer greater than or equal to 1, so it is maximum for n=1 or:

$|\frac{2\lambda \sin(\lambda \pi )}{\lambda ^2-n^2}|<
\frac{2(1+.5) }{1+.25}<2.4<3$

You could answer part 1 by assuming $\lambda < n-0.5$ and following the same procedure, then show that D is increasing on the range $n-.5<\lambda\leq n$ and therefore is less than D at n = lambda.

4. i think you made a mistake on the third line because: $\lambda ^2-n^2>n+0.25\Rightarrow\frac{2}{\lambda ^2-n^2}<\frac{2}{n+0.25}$ and $\lambda >n+0.5$

5. Not solved - inequality problem

Oops sorry. If you find a solution can you please post it, this is really bugging me!