True or False

Calculus Level 3

True or False?

$\int_0^{\infty} \frac{\cos x}{1+x} \, dx = \int_0^{\infty} \frac{\sin x}{(1+x)^2} \, dx$

False True

1 solution

Hana Wehbi
Dec 17, 2016

We can prove this by integration by parts by letting $\large u= \frac{1}{1+x}\implies du=\frac{-1}{(1+x)^2}dx$

and $\large dv= \cos(x)dx \implies v=\sin(x)$ . Then

$\int_0^{\infty} udv= uv\Bigg|_0^{\infty}- \int_0^{\infty} vdu$

$\int_0^{\infty}\frac{\cos(x)}{1+x} dx= \frac{\sin(x)}{1+x}\Bigg|_0^{\infty} + \int_0^{\infty} \frac{sin(x)}{(1+x)^2}dx$

$\int_0^{\infty}\frac{\cos(x)}{1+x} dx = 0 + \int_0^{\infty}\frac{\sin(x)}{(1+x)^2}dx$

You still have to show that the integral converges, otherwise, you could have $\infty = \infty$ .

Pi Han Goh - 4 years, 6 months ago

Yes, you are right. One of these integrals converges absolutely, but the other does not. I will add this to my solution. Thank you.

Hana Wehbi - 4 years, 5 months ago