Hot Integral - 6

Calculus Level 5

I n ( x ) = 2 n π ∫ − ∞ ∞ ( x + i t ) n e t 2 d t \qquad \qquad\displaystyle I_n(x) = \frac{2^n}{\sqrt{\pi}}\int\limits_{-\infty}^{\infty}\frac{(x+it)^n}{e^{t^2}} dt

Then, the following summation can be expressed as:

∑ n = 0 ∞ I n ( x ) 2 n ⌊ n 2 ⌋ ! = a + b x a c d e p a x q \displaystyle \sum_{n=0}^{\infty} \frac{I_n(x)2^n}{\lfloor \frac{n}{2}\rfloor !} = \frac{a+bx}{a^{\frac{c}{d}}}e^{\frac{p}{a}x^q}

Calculate a + b + c + d + p + q a+b+c+d+p+q .

Details and Assumptions

1) i = − 1 i=\sqrt{-1}

2) a , b , c , d , p , q a,b,c,d,p,q are positive integers.

3) gcd ⁡ ( c , d ) = gcd ⁡ ( p , a ) = 1 \gcd(c,d)=\gcd(p,a)=1

4)Sign ! ! indicates factorial. For e.g. 3 ! = 3 × 2 × 1 3! = 3 \times 2 \times 1

5) ⌊ . ⌋ \lfloor . \rfloor represents the floor function.

This is a part of Hot Integrals


The answer is 44.

Aman Rajput by Aman Rajput , 25, India

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1 solution

Kartik Sharma
Aug 12, 2015

Again a nice problem!

I'll straightaway put the integral into the sum. That is a valid step of course.

∑ n = 0 ∞ 2 n π ∫ − ∞ ∞ ( x + i t ) n e t 2 d t 2 n ⌊ n 2 ⌋ ! \displaystyle \sum_{n=0}^{\infty}{\frac{\displaystyle \frac{2^n}{\sqrt{\pi}}\int\limits_{-\infty}^{\infty}\frac{(x+it)^n}{e^{t^2}} dt \quad {2}^{n}}{\left \lfloor \frac{n}{2} \right \rfloor !}}

Now, we will sum under integral sign i.e. take the summation under the integral sign. That is again valid as integral is not taken w.r.t. n n .

1 π ∫ − ∞ ∞ e − t 2 ∑ n = 0 ∞ ( 4 ( x + i t ) ) n ⌊ n 2 ⌋ ! d t \displaystyle \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{{e}^{-{t}^{2}} \sum_{n=0}^{\infty}{\frac{{\left(4(x+it)\right)}^{n}}{\left \lfloor \frac{n}{2} \right \rfloor !}} \quad dt}

1 π ∫ − ∞ ∞ e − t 2 ( 4 ( x + i t ) + 1 ) e ( 4 x + 4 i t ) 2 d t \displaystyle \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{{e}^{-{t}^{2}} \left(4(x+it)+1\right){e}^{{\left(4x+4it\right)}^{2}} dt}

The above is true because -

∑ n = 0 ∞ a n ⌊ n 2 ⌋ ! = ∑ n = 0 ∞ a 2 n + a 2 n + 1 n ! = ( a + 1 ) e a 2 \displaystyle \sum_{n=0}^{\infty}{\frac{{a}^{n}}{\left \lfloor \frac{n}{2} \right \rfloor !}} = \sum_{n=0}^{\infty}{\frac{{a}^{2n} + {a}^{2n+1}}{n!}} = (a+1){e}^{{a}^{2}}

Hence, we will proceed with the integral now

1 π ∫ − ∞ ∞ ( 4 ( x + i t ) + 1 ) e − 17 t 2 + 32 x i t + 16 x 2 d t \displaystyle \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{ \left(4(x+it)+1\right){e}^{-17{t}^{2}+ 32xit + 16{x}^{2}} dt}

For computing this integral, we will first find some general integrals.

1. ∫ − ∞ ∞ e − a x 2 + b x + c d x = π a e b 2 4 a + c \displaystyle 1. \quad \int\limits_{-\infty}^{\infty}{{e}^{-a{x}^{2}+ bx + c} dx} = \displaystyle \sqrt{\frac{\pi}{a}} {e}^{\frac{{b}^{2}}{4a}+c}

Proof: ∫ − ∞ ∞ e − a x 2 + b x + c d x \displaystyle \text{Proof:} \quad \int\limits_{-\infty}^{\infty}{{e}^{-a{x}^{2}+ bx + c} dx}

= e b 2 4 a + c ∫ − ∞ ∞ e − a x 2 + b x − b 2 4 a d x \displaystyle = {e}^{\frac{{b}^{2}}{4a}+c} \int\limits_{-\infty}^{\infty}{{e}^{-a{x}^{2}+ bx -\frac{{b}^{2}}{4a}} dx}

= e b 2 4 a + c ∫ − ∞ ∞ e − ( a x − b 2 a ) 2 d x \displaystyle = {e}^{\frac{{b}^{2}}{4a}+c} \int\limits_{-\infty}^{\infty}{{e}^{-{\left(\sqrt{a}x - \frac{b}{2\sqrt{a}}\right)}^{2}} dx}

Substituting u = a x − b 2 a u = \sqrt{a}x - \frac{b}{2\sqrt{a}} and using Gaussian integral,

= π a e b 2 4 a + c \displaystyle = \sqrt{\frac{\pi}{a}} {e}^{\frac{{b}^{2}}{4a}+c}

2. ∫ − ∞ ∞ x e − a x 2 + b x + c d x = b e b 2 4 a + c 2 a π a \displaystyle 2. \quad \int\limits_{-\infty}^{\infty}{x{e}^{-a{x}^{2}+ bx + c} dx} = \frac{b{e}^{\frac{{b}^{2}}{4a}+c}}{2a}\sqrt{\frac{\pi}{a}}

Proof: \displaystyle \text{Proof:}

Moving ahead as we did in the previous proof (using quadratic substitution, u- substitution etc.), we'd simply get

e b 2 4 a a ∫ − ∞ ∞ ( u + b 2 a ) e − u 2 d u \displaystyle \frac{{e}^{\frac{{b}^{2}}{4a}}}{a} \int\limits_{-\infty}^{\infty}{\left(u+\frac{b}{2\sqrt{a}}\right){e}^{-{u}^{2}} du}

e b 2 4 a a ( ∫ − ∞ ∞ u e − u 2 d u + b 2 a ∫ − ∞ ∞ e − u 2 d u ) \displaystyle \frac{{e}^{\frac{{b}^{2}}{4a}}}{a} \left(\int\limits_{-\infty}^{\infty}{u{e}^{-{u}^{2}} du} + \frac{b}{2\sqrt{a}} \int\limits_{-\infty}^{\infty}{{e}^{-{u}^{2}} du}\right)

Now, the first integral will get equal to 0 0 since it is an odd function. And we are left with only the 2nd integral which is purely Gaussian. So, we'll get

b e b 2 4 a + c 2 a π a \displaystyle \frac{b{e}^{\frac{{b}^{2}}{4a}+c}}{2a}\sqrt{\frac{\pi}{a}}

That's all what we require to compute our integral. Let's get back to business.

This was our integral -

1 π ∫ − ∞ ∞ ( 4 ( x + i t ) + 1 ) e − 17 t 2 + 32 x i t + 16 x 2 d t \displaystyle \frac{1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{ \left(4(x+it)+1\right){e}^{-17{t}^{2}+ 32xit + 16{x}^{2}} dt}

Manipulating it,

4 x + 1 π ∫ − ∞ ∞ e − 17 t 2 + 32 x i t + 16 x 2 d t + 4 i π ∫ − ∞ ∞ t e − 17 t 2 + 32 x i t + 16 x 2 d t \displaystyle \frac{4x + 1}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{{e}^{-17{t}^{2}+ 32xit + 16{x}^{2}} dt} + \frac{4i}{\sqrt{\pi}} \int\limits_{-\infty}^{\infty}{t{e}^{-17{t}^{2}+ 32xit + 16{x}^{2}} dt}

Now, using the results we derived above,

= ( 4 x + 1 ) e − 32 2 x 2 4 × 17 + 16 x 2 17 − 64 x e − 32 2 x 2 4 × 17 + 16 x 2 17 17 \displaystyle = \frac{(4x+1){e}^{-\frac{{32}^{2}{x}^{2}}{4 \times 17} + 16{x}^{2}}}{\sqrt{17}} - \frac{64x{e}^{-\frac{{32}^{2}{x}^{2}}{4 \times 17} + 16{x}^{2}}}{17\sqrt{17}}

which on simplification gives -

( 4 x + 17 ) e 16 17 x 2 17 3 2 \displaystyle \boxed{\frac{\left(4x+17\right){e}^{\frac{16}{17}{x}^{2}}}{{17}^{\frac{3}{2}}}}

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