We will use integration by parts (twice):

`int u dv=uv-int v du`

`int_0^infty e^-x cos x dx=|[u=e^-x,dv=cos x dx],[du=-e^-x dx,v=sin x]|=`

`e^-x sin x+int_0^infty e^-x sin x dx=|[u=e^-x,dv=sin x dx],[du=-e^-x dx,v=-cos x]|=`

`(e^-x sin x-e^-x cos x)|_0^infty-int_0^inftye^-x cos x dx`

We can see that the end integral is equal to the starting one hence we can solve equation

`int_0^infty e^-x cos x dx=(e^-x sin x-e^-x cos x)|_0^infty-int_0^inftye^-x cos x dx`

`2int_0^infty e^-x cos x dx=(e^-x sin x-e^-x cos x)|_0^infty`

`int_0^infty e^-x cos x dx=1/2(e^-x sin x-e^-x cos x)|_0^infty=`

`1/2[lim_(x to infty) (e^-x sin x-e^-x cos x)-e^0 sin0+e^0 cos0]=`

`1/2(0-0-0+1)=1/2`

As we can see **the integral converges** to `1/2.`

The image below shows the graph of the function and area between it and `x`-axis corresponding to the integral. We can see that the function converges to zero and it does so "very fast" implying that the integral should probably converge, which we have shown in the above calculation.

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