Home page for accesible maths 1 1 Further Integration

Style control - access keys in brackets

Font (2 3) - + Letter spacing (4 5) - + Word spacing (6 7) - + Line spacing (8 9) - +

1.43 Integrals over infinite ranges

Example.

1.43.1 The integral ∫0∞cos⁡x⁢d⁢x\int_{0}^{\infty}\cos x\,dx diverges.

To see this, we have ∫0Rcos⁡x⁢d⁢x=[sin⁡x]0R=sin⁡R\int_{0}^{R}\cos x\,dx=[\sin x]_{0}^{R}=\sin R. Since sin⁡R\sin R doesn’t converge to a fixed value as R→∞R\rightarrow\infty, the integral diverges.

Example.

1.43.2 The following improper integral converges: ∫1∞1x⁢(x+1)⁢d⁢x=log⁡2.\int_{1}^{\infty}{{1}\over{x(x+1)}}\,dx=\log 2.

Solution. Using partial fractions, we have 1x⁢(x+1)=1x-1x+1,\frac{1}{x(x+1)}=\frac{1}{x}-\frac{1}{x+1}, so ∫1R1x⁢(x+1)⁢d⁢x=\int_{1}^{R}\frac{1}{x(x+1)}dx= [log⁡x-log⁡(x+1)]1R=[log⁡xx+1]1R[\log x-\log(x+1)]^{R}_{1}=[\log\frac{x}{x+1}]^{R}_{1}.

Now log⁡RR+1=log⁡11+1R→log⁡1=0\log\frac{R}{R+1}=\log\frac{1}{1+\frac{1}{R}}\rightarrow\log 1=0 as R→∞R\rightarrow\infty, so the integral converges to -log⁡12=log⁡2-\log\frac{1}{2}=\log 2.