Question

For what values of \(a.\)is \(\int_0^\infty {{e^{ax}}} cosxdx\)convergent? Evaluate the integral for those values of \(a.\)

Short answer

The integral will converge only if \(a < 0\), \(\int_0^\infty {{e^{ax}}} \cos xdx = - \frac{a}{{1 + {a^2}}}\)

Step-by-step solution

Step 1: Given Integration

Do integration by parts

\(\int {{e^{ax}}} \cos xdx = {e^{ax}}\int {\cos } xdx - \int {\left( {\frac{{d\left( {{e^{ax}}} \right)}}{{dx}}\int {\cos } xdx} \right)} dx{\rm{ }}\)

\(\int {{e^{ax}}} \cos xdx = {e^{ax}}\sin x - a\int {{e^{ax}}} \sin xdx \to {\rm{ }}\)Equation 1

Step 2: Reverse order integration part

Do integration by parts in the reverse order

\(\int {{e^{ax}}} \cos xdx = \cos x\int {{e^{ax}}} dx - \int {\left( {\frac{{d(\cos x)}}{{dx}}\int {{e^{ax}}} dx} \right)} dx{\rm{ }}\)

\(\int {{e^{ax}}} \cos xdx = \frac{1}{a} \cdot \cos x \cdot {e^{ax}} + \frac{1}{a}\int {\sin } x \cdot {e^{ax}}dx \to {\rm{ }}\)Equation 2

Step 3: Multiply and Divide both equation

Multiply Equation\(2\)by\({a^2}\)and add to Equation\(1\)

\(\left( {1 + {a^2}} \right)\int {{e^{ax}}} \cos xdx = a\cos x \cdot {e^{ax}} + {e^{ax}}\sin x\)

Divide both sides by\(1 + {a^2}\)

\(\int {{e^{ax}}} \cos xdx = \frac{{{e^{ax}}(a\cos x + \sin x)}}{{1 + {a^2}}}\)

Step 4: Final Proof

Therefore

\(\int_0^\infty {{e^{ax}}} \cos xdx = \left( {\frac{{{e^{ax}}(a\cos x + \sin x)}}{{1 + {a^2}}}} \right)_0^\infty \)

Notice that the integral will converge only if\(a < 0\)

\(\int_0^\infty {{e^{ax}}} \cos xdx = - \frac{a}{{1 + {a^2}}}\)