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Question: If \({I_n} = \int\limits_0^\infty {{e^{ - x}}} {x^{n - 1}}dx,\) then \(\int\limits_0^\infty {{e^{ - ...

If In=∫0∞e−xxn−1dx,{I_n} = \int\limits_0^\infty {{e^{ - x}}} {x^{n - 1}}dx, then ∫0∞e−λxxn−1dx\int\limits_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} is equal to which of the following options?
A) λIn\lambda {I_n}
B) 1λIn\dfrac{1}{\lambda }{I_n}
C) Inλn\dfrac{{{I_n}}}{{{\lambda ^n}}}
D) λnIn{\lambda ^n}{I_n}

Explanation

Solution

In this question, we are given an equation in integration and we have been asked the value of another similar equation. Using the given equation, we have to find the value of another equation in terms of the given equation. Start by the equation which has λ\lambda in it and substitute λx\lambda x by some other variable. Then differentiate this and put in the equation with which we started. Simplify the equation and take λ\lambda common. After that, use the property of changing the variable directly and change it into ‘x’. The final equation will be equal to that given in the question. Substitute the value given in question and you will get the final answer.

Complete step-by-step answer:
We are given an equation In=∫0∞e−xxn−1dx{I_n} = \int\limits_0^\infty {{e^{ - x}}} {x^{n - 1}}dx and we have been asked to find the value of ∫0∞e−λxxn−1dx\int\limits_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} . We will start by taking ∫0∞e−λxxn−1dx\int\limits_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} .
⇒∫0∞e−λxxn−1dx\Rightarrow \int\limits_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} …………..…. (1)
We will substitute λx=t\lambda x = t
Differentiating both the sides with respect to x,
⇒λ=dtdx\Rightarrow \lambda = \dfrac{{dt}}{{dx}}
⇒dx=dtλ\Rightarrow dx = \dfrac{{dt}}{\lambda }
Now we will substitute this in equation (1) to bring it in terms of ‘t’,
⇒∫0∞e−t(tλ)n−1dtλ\Rightarrow \int\limits_0^\infty {{e^{ - t}}{{\left( {\dfrac{t}{\lambda }} \right)}^{n - 1}}\dfrac{{dt}}{\lambda }}
On simplifying we will get,
⇒∫0∞e−ttn−1λn−1dtλ\Rightarrow \int\limits_0^\infty {{e^{ - t}}\dfrac{{{t^{n - 1}}}}{{{\lambda ^{n - 1}}}}\dfrac{{dt}}{\lambda }}
Now, we will simplify the denominator,
⇒∫0∞e−ttn−1λn−1+1dt\Rightarrow \int\limits_0^\infty {{e^{ - t}}\dfrac{{{t^{n - 1}}}}{{{\lambda ^{n - 1 + 1}}}}dt} …. (Using property ab×ac=ab+c{a^b} \times {a^c} = {a^{b + c}})
⇒∫0∞e−ttn−1λndt\Rightarrow \int\limits_0^\infty {{e^{ - t}}\dfrac{{{t^{n - 1}}}}{{{\lambda ^n}}}dt}
Now, we will take out λn{\lambda ^n} common,
⇒1λn∫0∞e−ttn−1dt\Rightarrow \dfrac{1}{{{\lambda ^n}}}\int\limits_0^\infty {{e^{ - t}}{t^{n - 1}}dt}
We will again change this equation in terms of ‘x’. As per a property of integration, we can change the variables. So, we will change ‘t’ into ‘x’.
⇒1λn∫0∞e−xxn−1dx\Rightarrow \dfrac{1}{{{\lambda ^n}}}\int\limits_0^\infty {{e^{ - x}}{x^{n - 1}}dx} ………..…. (2)
Now, if you see clearly, you will notice that this same equation is given in the question and it is equal to In{I_n}.
Putting, In=∫0∞e−xxn−1dx{I_n} = \int\limits_0^\infty {{e^{ - x}}} {x^{n - 1}}dx in equation (2),
⇒1λnIn\Rightarrow \dfrac{1}{{{\lambda ^n}}}{I_n}
Therefore, ∫0∞e−λxxn−1dx\int\limits_0^\infty {{e^{ - \lambda x}}{x^{n - 1}}dx} = 1λnIn\dfrac{1}{{{\lambda ^n}}}{I_n}

Option C is the correct answer.

Note: We can observe that in mathematics, the exponential integral is the special function on the complex plane. It is defined as one particular definite integral of the ratio between an exponential function and its argument.