Evaluation of $\int^{\infty}_{0}e^{-ax}(\cos x)^ndx$

Evaluation of $$\int^{\infty}_{0}e^{-ax}(\cos x)^ndx,a>0$$

$\bf{My\; Try::}$ Let $$I = \int^{\infty}_{0}e^{-ax}(\cos x)^ndx\;,$$

Now Using euler substution $$\cos x = \frac{e^{-ix}+e^{-ix}}{2}$$

So $$I = \int^{\infty}_{0}e^{-ax}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^ndx$$

Now Use Binomial expansion,But i did not understand how can i solve after that, Help required

Thanks

• you can integrate $e^{-ax}cos(bx)$ by using integration by parts twice, then rearranging in terms of your original integral, which will then appear on the left and right hand sides. you could use the $(\frac{e^{-ix}+e^{-ix}}{2})^n$ expansion to get an expression for $(cos(x))^n$ in terms of a cos bx with various a and b – Cato Oct 11 '16 at 9:30

2 Answers

Hint: $$\int_0^\infty e^{(-a+in)x}dx=\frac{1}{-a+in}e^{(-a+in)x}\Big|_0^\infty=\frac{1}{in-a}$$ Of course, assuming that $a>0$.

Let $I_{n}=\int_{0}^{\infty}e^{-ax}(cos(x))^ndx$

Then we know that $I_{0}=1/a$, and $I_{1}=-a/(1+a^2)$

Now, using integration by parts,

$I_n=\int_{0}^{\infty}e^{-ax}(cos(x))^{n-1}cos(x)dx$

$=e^{-ax}(cos(x))^{n-1} sin(x)|_{0}^{\infty} - \int_{0}^{\infty}[e^{-ax}.(-a)(cos(x))^{n-1}-e^{-ax}.(cos(x))^{n-2}.(n-1).sin(x)](\int cos(x) dx) dx$

$=0 + a\int _{0}^{\infty}e^{-ax}.(cos(x))^{n-1}.sin(x)dx + (n-1) \int_{0}^{\infty}e^{-ax}.(cos(x))^{n-2}(1-cos^{2}(x)) dx$

$=a\int _{0}^{\infty}e^{-ax}.(cos(x))^{n-1}.sin(x)dx + (n-1)(I_{n-2}-I_n)$

$=a[(e^{-ax}\int (cos(x))^{n-1}.sin(x)dx)|_{0}^{\infty} -\int_{0}^{\infty} (-a).e^{-ax}(\int (cos(x))^{n-1}.sin(x)dx)dx] + (n-1)(I_{n-2}-I_n)$

$= -a/n + (a^2/n)I_n + (n-1)(I_{n-2}-I_n)$

So we have

$I_n = -a/n + (a^2/n)I_n + (n-1)(I_{n-2}-I_n)$

which gives us

$((n^2-a^2)/n)I_n=(n-1)I_{n-2}-a/n$

This is a recursive relation and gives a formula for $I_n$ in $I_{0}$ for n even and $I_1$ for n odd.